datavorous/entrance-exam-dataset huggingface dataset:
The equation of the plane passing through the line of intersection of the planes r→·(i^+j^+k^)=1 and r→·(2i^+3j^-k^)+4=0 and parallel to the x-axis, is

Options:
A r→·(i^+3k^)+6=0
B r→·(i^-3k^)+6=0
C r→·(j^-3k^)-6=0
D r→·(j^-3k^)+6=0

The correct answer is:
r→·(j^-3k^)+6=0

Solution:
P1:r→·(i^+j^+k^)=1

⇒x+y+z-1=0

P2: r→·(2i^+3j^-k^)+4=0

⇒2x+3y-z+4=0

The equation of the plane passing through the line of intersection of the planes:P1+λP2=0

(1+2λ)x+(1+3λ)y+(1-λ)z=1-4λ...i
Parallel to x-axis

a1a2+b1b2+c1c2=0

(1+2λ)1+(1+3λ)0+(1−λ)0=0

λ=−12

1−32y+1+12z=1+2

3z−y=6

r→⋅(j^−3k^)+6=0

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2021 (27 Aug Shift 2)
The equation of the plane passing through the line of intersection of the planes $x+y+z=1$ and $2 x+3 y-z+4=0$ and parallel to $x$-axis is

Options:
A $y-3 z-6=0$
B $y-3 z+6=0$
C $y-z-1=0$
D $y-z+1=0$

The correct answer is:
$y-3 z+6=0$

Solution:
Equation of plane passing through intersection of given planes is, $(x+y+z-1)+\lambda(2 x+3 y-z+4)=0$
Plane (i) is parallel to $x$-axis, then $(1+2 \lambda) 1=0$
$\Rightarrow \lambda=-\frac{1}{2}$
Put the value of $\lambda$ in (i), we get $y-3 z+6=0$, which is the required plane

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The equation of the plane passing through the lines $\frac{x-4}{1}=\frac{y-3}{1}=\frac{z-2}{2}$ and $\frac{x-3}{1}=\frac{y-2}{-4}=\frac{z}{5}$ is

Options:
A $11 x-y-3 z=35$
B $11 x+y-3 z=35$
C $11 x-y+3 z=35$
D None of these

The correct answer is:
None of these

Solution:
$\begin{aligned}
& a(x-4)+b(y-3)+c(z-2)=0 \\
& \therefore a+b+2 c=0 \text { and } a-4 b+5 c=0 \\
& \frac{a}{5+8}=\frac{b}{2-5}=\frac{c}{-4-1}=k \\
& \frac{a}{13}=\frac{b}{-3}=\frac{c}{-5}=k
\end{aligned}$
Therefore, the required equation of plane is
$-13 x+3 y+5 z+33=0$

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The equation of the plane passing through the origin and perpendicular to the line $x=2 y=3 z$ is

Options:
A $6 x+3 y+2 z=0$
B $x+2 y+3 z=0$
C $3 x+2 y+z=0$
D None of these

The correct answer is:
$6 x+3 y+2 z=0$

Solution:
Obviously the line is $\frac{x}{6}=\frac{y}{3}=\frac{z}{2}$. So required plane is $6(x-0)+3(y-0)+2(z-0)=0$.

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The equation of the plane passing through the point 1, 2,-3 and perpendicular to the planes 3x+y-2z=5 and 2x-5y-z=7, is

Options:
A 11x+y+17z+38=0
B 3x-10y-2z+11=0
C 6x-5y+2z+10=0
D 6x-5y-2z-2=0

The correct answer is:
11x+y+17z+38=0

Solution:
Normal vector of the plane will be

i^j^k^31-22-5-1=-11i^-j^-17k^

So, the direction ratios of normal to the required plane are 11,1,17 and plane passes through 1, 2,-3. So, the equation of plane is

11x-1+1y-2+17z+3=0

⇒11x+y+17z+38=0

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2021 (24 Feb Shift 1)
The equation of the plane passing through the point \( (-1,3,2) \) and perpendicular to each of the planes \( x+2 y+3 z=5 \) and \( 3 x+3 y+z=0 \), is

Options:
A \( 7 x-8 y+3 z-25=0 \)
B \( 7 x-8 y+3 z+25=0 \)
C \( -7 x+8 y-3 z+5=0 \)
D \( 7 x-8 y-3 z+5=0 \)

The correct answer is:
\( 7 x-8 y+3 z+25=0 \)

Solution:
Given, equation of plane is passing through the point -1, 3, 2
∴ Ax+1+By-3+Cz-2=0 .....(i)
Since plane (i) is perpendicular to each of the planes x+2y+3z=5 and 3x+3y+z=0
So, A+2B+3C=0 and 3A+3B+C=0
∴A2-9=B9-1=C3-6=K
A=-7K, B=8K, C=-3K
Put the values of A, B and C in (i)
we get, 7x-8y+3z+25=0 , which is the required equation of the plane.

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The equation of the plane passing through the point of intersection of the lines x-13=y-21=z-32, x-31=y-12=z-23 and perpendicular to the line x-22=y-33=z-21 is P=0. If the distance of the point 1,1,3 from P=0 is k units, then the value of k22 is equal to

Options:

The correct answer is:
7

Solution:
Any general point on the line x-13=y-21=z-32 is 3Î»+1,Î»+2,2Î»+3

For the point of intersection, this point must satisfy the line x-31=y-12=z-23
â3Î»-21=Î»+12=2Î»+13âÎ»=1
So, point of intersection is 4,3,5
The required equation of plane is
2x-4+3y-3+1z-5=0
â2x+3y+z-22=0
Hence, the distance of the plane from 1,1,3 =2+3+3-2214=14=k

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The equation of the plane passing through the points $(0,1,2)$ and $(-1,0,3)$ and perpendicular to the plane $2 x+3 y+z=5$ is

Options:
A $3 x-4 y+18 z+32=0$
B $3 x+4 y-18 z+32=0$
C $4 x+3 y-17 z+31=0$
D $4 x-3 y+z+1=0$

The correct answer is:
$4 x-3 y+z+1=0$

Solution:
Equation of any plane passing through $(0,1,2)$ is
$a(x-0)+b(y-1)+c(z-2)=0$
Plane (i) passes through $(-1,0,3)$, then
$\begin{aligned}
& a(-1-0)+b(0-1)+c(3-2)-0 \\
& \Rightarrow-a-b+c=0 \Rightarrow a+b-c=0
\end{aligned}$
Plane (i) is perpendicular to the plane $2 x+3 y+z=5$, then $2 a+3 b+c=0$
Solving (ii) and (iii), we get $a=-4 k, b=3 k, c=-k$
Putting these values in (i),
$\begin{aligned}
& -4 k(x)+3 k(y-1)-k(z-2)=0 \\
\Rightarrow & -4 x+3 y-3-z+2=0 \\
\Rightarrow & -4 x+3 y-z-1=0 \Rightarrow 4 x-3 y+z+1=0 .
\end{aligned}$

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The equation of the plane passing through the points
$(3,2,2)$ and $(1,0,-1)$ and parallel to the line $\frac{x-1}{2}=\frac{y-1}{-2}=\frac{z-2}{3}$, is

Options:
A $4 x-y-2 z+6=0$
B $4 x-y+2 z+6=0$
C $4 x-y-2 z-6=0$
D None of these

The correct answer is:
None of these

Solution:
Equation of plane passing through the point $(1,0,-1)$ is,
$a(x-1)+b(y-0)+c(z+1)=0$
Also, plane (i) is passing through $(3,2,2)$
$\therefore \quad a(3-1)+b(2-0)+c(2+1)=0$
or $2 a+2 b+3 c=0$
Plane (i) is also parallel to the line $\frac{x-1}{2}=\frac{y-1}{-2}=\frac{z-2}{3}$
$\therefore 2 a-2 b+3 c=0$
From (i) and (ii), $\frac{a}{-3}=\frac{b}{0}=\frac{c}{2}$
Therefore, the required plane is,
$\begin{aligned}
& -3(x-1)+0(y-0)+2(z+1)=0 \\
& -3 x+2 z+5=0
\end{aligned}$

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The equation of the plane passing through the points \( (3,2,2) \) and \( (1,0,-1) \) and parallel to the line \( \frac{x-1}{2}=\frac{y-1}{-2}=\frac{z-2}{3} \), is

Options:
A \( 4 x-y-2 z+6=0 \)
B \( 4 x-y+2 z+6=0 \)
C \( 4 x-y-2 z-6=0 \)
D None of these

The correct answer is:
None of these

Solution:
Equation of plane passing through the point 1,0,-1 is,
ax-1+by-0+cz+1=0 ......(i)
Also, plane (i) is passing through 3, 2, 2
∴ a 3-1+b 2-0+c 2+1=0
or 2a+2b+3c=0 .....(i)
Plane (i) is also parallel to the line x-12=y-1-2=z-23
∴2a-2b+3c=0 .....(ii)
From (i) and (ii), a-3=b0=c2
Therefore, the required plane is,
-3 x-1+0 y-0+2 z+1=0
or -3x+2z+5=0 .

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The equation of the plane through (1,2,-3) and (2,-2,1) and parallel to $X$ -axis is

Options:
A $y-z+1=0$
B $y-z-1=0$
C $y+z-1=0$
D $y+z+1=0$

The correct answer is:
$y+z+1=0$

Solution:
Equation of the plane will be $\left|\begin{array}{ccc}x-1 & y-2 & z+3 \\ 2-1 & -2-2 & 1-(-3) \\ 1 & 0 & 0\end{array}\right|=0$
$\Rightarrow\left|\begin{array}{ccc}x-1 & y-2 & z+3 \\ 1 & -4 & 4 \\ 1 & 0 & 0\end{array}\right|=0$
$\Rightarrow \quad 1[4(y-2)+4(z+3)]=0$
[Expanding along $\left.R_{j}\right]$
$\Rightarrow \quad 4 y-8+4 z+12=0$
$\Rightarrow \quad 4 y+4 z+4=0$
$\Rightarrow \quad y+z+1=0$

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The equation of the plane through the intersection of the planes $x+y+z=1$ and $2 x+3 y-z+4$ and parallel to the $x$-axis is

Options:
A $y+3 z+6=0$
B $y+3 z-6=0$
C $y-3 z+6=0$
D $y-3 z-6=0$

The correct answer is:
$y-3 z+6=0$

Solution:
$\quad(2 x+3 y-z+4)+\lambda(x + y+z-1)=0$
$\overrightarrow{\mathrm{n}}=(2+\lambda) \hat{\mathrm{i}}+(3+\lambda) \hat{\mathrm{j}}+(-1+\lambda) \hat{\mathrm{k}}$
$\overrightarrow{\mathrm{n}} \cdot \overrightarrow{\mathrm{i}}=0 \Rightarrow \lambda=-2$
$\Rightarrow \mathrm{y}-3 z+6=0$

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The equation of the plane through the point $(2,-1,-3)$ and parallel to the lines $\frac{x-1}{3}=\frac{y+2}{2}=\frac{z}{-4}$ and $\frac{x}{2}=\frac{y-1}{-3}=\frac{z-2}{2}$ is

Options:
A $8 x+14 y+13 z+37=0$
B $8 x-14 y+13 z+37=0$
C $8 x+14 y-13 z+37=0$
D $8 x+14 y+13 z-37=0$

The correct answer is:
$8 x+14 y+13 z+37=0$

Solution:
Equation of plane passing through the point $(2,-1,-3)$ is, $A(x-2)+B(y+1)+C(z+3)=0$
Also, $3 A+2 B-4 C=0$ and $2 A-3 B+2 C=0$
$\therefore \frac{A}{-8}=\frac{B}{-14}=\frac{C}{-13}=k$
, (Let)
So, $A=-8 k, B=-14 k, C=-13 k$
Equation of required plane is,
$-k[8(x-2)+14(y+1)+13(z+3)]=0$
i.e., $8 x+14 y+13 z+37=0$.

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The equation of the plane through the point \( (0,-4,-6) \) and \( (-2,9,3) \) and perpendicular to the plane \( x-4 y-2 z=8 \) is

Options:
A \( 3 x+3 y-2 z=0 \)
B \( x-2 y+\mathrm{z}=2 \)
C \( 2 x+y-\mathrm{z}=2 \)
D \( 5 x-3 y+2 z=0 \)

The correct answer is:
\( 2 x+y-\mathrm{z}=2 \)

Solution:
Equation of plane through (0,-4,-6) is
ax-0+by+4+cz+6=0 ...(i)
Point (-2,9,3) lies on Eq. (i), then
-2a+13b+9c=0 ...(ii)
Also required plane is perpendicular to x-4y-2z=8
∴a-4b-2c=0 ...(iii)
From Eqs. (i), (ii), (iii) we get
xy+4z+6-21391-4-2=0

⇒ x-26+36-y+44-9x+68-13=0
⇒2x+y-z-2=0

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The equation of the plane through the point \( (1,2,-3) \) which is parallel to the plane \( 3 x-5 y+2 z=11 \) is given by

Options:
A \( 3 x-5 y+2 z-13=0 \)
B \( 5 x-3 y+2 z+13=0 \)
C \( 3 x-2 y+5 z+13=0 \)
D \( 3 x-5 y+2 z+13=0 \)

The correct answer is:
\( 3 x-5 y+2 z+13=0 \)

Solution:
The equation of the plane parallel to 3 x-5 y+2 z=11 is 3 x-5 y+2 z=k as they are proportional.

The equation of the above plane passing through the point (1, 2, -3) is given by

(3×1)-(5×2)+(2×-3)=kk=3-10-6k=-13

Now, we can substitute the value of k in the first equation to get the equation of the required plane, i.e.

3x-5y+2z=-13 or 3x-5y+2z+13=0

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The equation of the plane through the points 2,-1,0, 3,-4,5 and parallel to a line with direction cosines proportional to 2,3 and 4 is 9x-2y-3z=k , then the value of k is

Options:
A 20
B -20
C 10
D -10

The correct answer is:
20

Solution:
Equation of the plane through 2,-1,0 is
ax-2+by+1+cz-0=0
It also passes through 3,-4,5, then
a-3b+5c=0... (i)
Given plane is parallel to a line with direction cosines proportional to 2,3,4, so
2a+3b+4c=0... (ii)
Solving equations (i) and (ii),
a-12-15=b10-4=c3+6
a-27=b6=c9
a-9=b2=c3=Î»
a=-9Î»,b=2Î»,c=3Î»
So, the equation of the plane is
-9Î»x-2+2Î»y+1+3Î»z-0=0
-9x+18+2y+2+3z=0
-9x+2y+3z=-20
9x-2y-3z=20
On comparison, k=20

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The equation of the plane through the points \( (1,2,3),(-1,4,2) \&(3,1,1) \) is

Options:
A \( 5 x+y+12 z-23=0 \)
B \( 5 x+6 y+2 z-23=0 \)
C \( x+6 y+2 z-13=0 \)
D \( x+y+z-13=0 \)

The correct answer is:
\( 5 x+6 y+2 z-23=0 \)

Solution:
Equation of plane through 1, 2, 3 is
ax-1+by-2+cz-3=0 ....(i)
∵ It passes through -1, 4,2 and 3, 1, 1
∴-2a+2b-c=0 and 2a-b-2c=0
⇒ a-5=b-6=c-2
∴ Equation of plane is
-5x-6y-2z+5+12+6=0
⇒ 5x+6y+2z-23=0

Alternate Solution

Equation plane is
x-x1y-y1z-z1x2-x1y2-y1z2-z1x3-x1y3-y1z3-z1=0
⇒ x-1y-2z-3-22-12-1-2=0
⇒ x-1-4-1-y-24+2+z-32-4=0
⇒ -5x+5-6y+12-2z+6=0
⇒ 5x+6y+2z-23=0

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The equation of the plane through the three points $(1,1,1),(1,-1,1)$ and $(-7,-3,-5)$, is

Options:
A $3 x-4 z+1=0$
B $3 x-4 y+1=0$
C $3 x+4 y+1=0$
D None of these

The correct answer is:
$3 x-4 z+1=0$

Solution:
Plane passing through $(1,1,1)$ is $a(x-1)+b(y-1)+c(z-1)=0$
It also passes through $(1,-1,1)$ and $(-7,-3,-5)$, then
$b=0, a=-\frac{3}{4} c$
Hence the required equation is $3 x-4 z+1=0$
Trick : Since the plane $3 x-4 z+1=0$ passes through given three points.

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The equation of the plane which bisects the angle between the planes $3 x-6 y+2 z+5=0$ and $4 x-12 y+3 z-3=0$ which contains the origin is

Options:
A $33 x-13 y+32 z+45=0$
B $x-3 y+z-5=0$
C $33 x+13 y+32 z+45=0$
D None of these

The correct answer is:
None of these

Solution:
Equation of plane bisecting the angle containing origin is (making constant term of same sign)
$\begin{aligned}
& \frac{-3 x+6 y-2 z-5}{\sqrt{3^2+6^2+2^2}}=+\left[\frac{4 x-12 y+3 z-3}{\sqrt{4^2+12^2+3^2}}\right] \\
& \text { or } \frac{-3 x+6 y-2 z-5}{7}=\frac{4 x-12 y+3 z-3}{13}
\end{aligned}$
or
$67 x-162 y+47 z+44=0$

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The equation of the plane which bisects the line joining $(2,3,4)$ and $(6,7,8)$ is

Options:
A $x+y+z-15=0$
B $x-y+z-15=0$
C $x-y-z-15=0$
D $x+y+z+15=0$

The correct answer is:
$x+y+z-15=0$

Solution:
Mid-point of $(2,3,4)$ and $(6,7,8)$ is $(4,5,6)$. This lies on $x+y+z-15=0$. Hence this is the required plane.

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The equation of the plane which bisects the line joining the points $(-1,2,3)$ and $(3,-5,6)$ at right angle, is

Options:
A $4 x-7 y-3 z=8$
B $4 x+2 y-3 z=28$
C $4 x-7 y+3 z=28$
D $4 x-7 y-3 z=28$

The correct answer is:
$4 x-7 y+3 z=28$

Solution:
According to question, $4 x-7 y+3 z=k$
Also, plane (i) passes through $\left(\frac{-1+3}{2}, \frac{2-5}{2}, \frac{3+6}{2}\right)$, then $4+\frac{21}{2}+\frac{27}{2}=k \Rightarrow k=28$
Therefore, required equation is $4 x-7 y+3 z=28$

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The equation of the plane which contains the y-axis and passes through the point 1, 2, 3 is:

Options:
A x+3z=10
B x+3z=0
C 3x+z=6
D 3x-z=0

The correct answer is:
3x-z=0

Solution:
n→=j^×(i^+2j^+3k^)

=3i^+0j^-k^

So, (3)(x-1)+0(y-2)+(-1)(z-3)=0

⇒3x-z=0

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2021 (17 Mar Shift 1)
The equation of the plane which is parallel to the plane $x-2 y+2 z=5$ and whose distance from the point $(1,2,3)$ is 1 , is

Options:
A $x-2 y+2 z=3$
B $x-2 y+2 z+3=0$
C $x-2 y+2 z=6$
D $x-2 y+2 z+6=0$

The correct answer is:
$x-2 y+2 z=6$

Solution:
Equation of plane parallel to $x-2 y+2 z=5$ is $x-2 y+2 z+k=0$ ..(i)
Now, according to question,
$\frac{1-4+6+k}{\sqrt{9}}= \pm 1$
or
$\begin{aligned}
& \text { or } k+3= \pm 3 \Rightarrow k=0 \text { or }-6 \\
& \therefore x-2 y+2 z-6=0 \text { or } x-2 y+2 z=6
\end{aligned}$

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The equation of the plane which is parallel to $x y$-plane and cuts intercept of length 3 from the $z$-axis is

Options:
A $x=3$
B $y=3$
C $z=3$
D $x+y+z=3$

The correct answer is:
$z=3$

Solution:
Equation of a plane parallel to $x y$-plane is $z=k$
$\therefore z=3$

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The equation of the plane which is parallel to $y$-axis and cuts off intercepts of length 2 and 3 from $x$-axis and $z$-axis is

Options:
A $3 x+2 z=1$
B $3 x+2 z=6$
C $2 x+3 z=6$
D $3 x+2 z=0$

The correct answer is:
$3 x+2 z=6$

Solution:
Equation of plane parallel to $y$-axis is, $a x+b z+1=0$
Also
$\begin{aligned}
& \quad 2 a+1=0 \Rightarrow a=-\frac{1}{2} \text { and } 3 b+1=0 \Rightarrow b=-\frac{1}{3} \\
& \text { Also } \\
& \therefore 3 x+2 z=6
\end{aligned}$
Alter : Equation of plane $\frac{x}{2}+\frac{z}{3}=1 \Rightarrow 3 x+2 z=6$

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The equation of the plane which passes through $(2,-3,1)$ and is normal to the line joining the points $(3,4,-1)$ and $(2,-1,5)$, is given by

Options:
A $x+5 y-6 z+19=0$
B $x-5 y+6 z-19=0$
C $x+5 y+6 z+19=0$
D $x-5 y-6 z-19=0$

The correct answer is:
$x+5 y-6 z+19=0$

Solution:
The equation of plane is
$$
\begin{array}{lr}
(x-2)+5(y+3)-6(z-1)=0 \\
\Rightarrow \quad x+5 y-6 z+19=0
\end{array}
$$

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The equation of the plane which passes through the point of intersection of x-13=y-21=z-32 and x-31=y-12=z-23 and perpendicular to 4i^+3j^+5k^, is

Options:
A 4x+3y+5z=50
B 4x+3y+5z=25
C 4x+3y+5z=10
D 4x-3y+5z=50

The correct answer is:
4x+3y+5z=50

Solution:
Any general point on the line x-13=y-21=z-32 is

3λ+1,λ+2,2λ+3
This must satisfy the given 2nd line

⇒3λ-21=λ+12=2λ+13
⇒λ=1
So, the point of intersection of the two lines are 4,3,5
Equation of the required plane is

4x-4+3y-3+5z-5=0
⇒4x-16+3y-9+5z-25=0
⇒4x+3y+5z-50=0

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The equation of the planes parallel to the plane x-2y+2z-3=0 which are at unit distance from the point 1, 2, 3 is ax+by+cz+d=0. If (b-d)=K(c-a), then the positive value of K is

Options:

The correct answer is:
4

Solution:
Let plane parallel to given plane is x-2y+2z+λ=0

Distance from 1, 2, 3=1

⇒|λ+3|5=1⇒λ=0,-6

On comparing with given equation

⇒a=1,b=-2,c=2,d=-6 or 0

b-d=4 or -2, c-a=1

⇒K=4 or -2

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2021 (18 Mar Shift 1)
The equation of the projection line of the line x+12=y+1-1=z+34 on the plane x+2y+z=6 is

Options:
A x-14=y-37=z-110
B x-1-4=y+37=z-110
C x-14=y-3-7=z+110
D x+34=y-27=z-7-10

The correct answer is:
x-14=y-3-7=z+110

Solution:
From the diagram,
Coordinates of the point A are 2Î»-1,-Î»-1,4Î»-3 (from the line)
It must satisfy the equation of the plane

2Î»-1-2Î»-2+4Î»-3=6âÎ»=3
So, coordinates of the point A are 5,-4,9
Point B â² must be the foot of perpendicular for point B
x+11=y+12=z+31=--1-2-3-61+4+1=2
x=1,y=3,z=-1

Direction ratios of <AB'>=<4,-7,10>
So, the equation of projection of line is x-14=y-3-7=z+110

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The equation of the resultant motion of the number of simple harmonic motions is Ec(1+Ksin⁡ω2t)sinω1t. The number of simple harmonic components is/are

Options:
A 1
B 2
C 3
D 4

The correct answer is:
3

Solution:
Y=Ec(1+Ksinâ¡Ï2t)sin Ï1t
Y=Ecsinâ¡Ï2t+Ec ksinâ¡Ï2sinâ¡Ï1t
Y=Ecsinâ¡Ï1t+EcK2cosâ¡Ï2+Ï1tEcK2Ï2-Ï1t
Thus there are three simple harmonic waves of frequency Ï1, Ï1+Ï2 and Ï1-Ï2

Topic: Physics, Oscillations, JEE Main
The equation of the smallest degree with real coefficients having $1+i$ as one of the root is

Options:
A $x^2+x+1=0$
B $x^2-2 x+2=0$
C $x^2+2 x+2=0$
D $x^2+2 x-2=0$

The correct answer is:
$x^2-2 x+2=0$

Solution:
Let $x=1+i \Leftrightarrow x-1=i$
$\Leftrightarrow(x-1)^2=i^2 \Leftrightarrow x^2-2 x+1=-1$
$\Leftrightarrow x^2-2 x+2=0$

Topic: Mathematics, Quadratic Equation, JEE Main
The equation of the straight line in the normal form which is parallel to the lines \( x+2 y+3=0 \) and \( x+2 y+8=0 \) and dividing the distance between these two lines in the ratio \( 1: 2 \) internally is

Options:
A \( x \cos a+y \sin a=\frac{10}{\sqrt{45}}, a=\tan ^{-1} \sqrt{2} \)
B \( x \cos a+y \sin a=\frac{14}{\sqrt{45}}, a=\pi+\tan ^{-1} 2 \)
C \( x \cos a+y \sin a=\frac{14}{\sqrt{45}}, a=\tan ^{-1} 2 \)
D \( x \cos a+y \sin a=\frac{10}{\sqrt{45}}, a=\pi+\tan ^{-1} \sqrt{2} \)

The correct answer is:
\( x \cos a+y \sin a=\frac{14}{\sqrt{45}}, a=\pi+\tan ^{-1} 2 \)

Solution:
We have,

x+2y+3=0 ...i

x+2y+8=0 ...ii

Required line is parallel to i & ii, therefore let the equation of required line be

x+2y+k=0 ...iii

Distance between i & iii is A=k-312+22=k-35.

Distance between ii & iii is

B=8-k12+22=8-k5

According to question,

AB=12

⇒k-35×58-k=12

⇒k-38-k=12

⇒k-38-k=±12

⇒2k-6=8-k or -2k+6=8-k

⇒k=143 or k=-2

Hence, equation of required line is

x+2y+143=0 or x+2y-2=0

So, the normal form of equation -x-2y=143 is -x5-25y=1435

⇒xcosα+ysinα=1445

where, α=π+tan-12

And, normal form of line x+2y=2 is

x5+25y=25

⇒xcosβ+ysinβ=25

where, β=tan-12

Topic: Mathematics, Straight Lines, JEE Main
The equation of the straight line passing through $(1,2,3)$ and perpendicular to the plane $x+2 y-5 z+9=0$ is

Options:
A $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{-5}$
B $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z+5}{3}$
C $\frac{x+1}{1}=\frac{y+2}{2}=\frac{z+3}{-5}$
D $\frac{x+1}{1}=\frac{y+2}{2}=\frac{z-5}{3}$

The correct answer is:
$\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{-5}$

Solution:
The line passes through point $(1,2,3)$ is $\frac{x-1}{a}=\frac{y-2}{b}=\frac{z-3}{c}$ $x+2 y-5 z+9=0$, therefore the line must be $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{-5}$ because
$\begin{aligned}
& \sin \theta=\frac{1 \cdot 1+2 \cdot 2+(-5)(-5)}{\sqrt{1^2+2^2+5^2} \cdot \sqrt{1^2+2^2+5^2}}=1 \\
& \Rightarrow \theta=90^{\circ} .
\end{aligned}$

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The equation of the straight line passing through the point $(3,2)$ and inclined at angle of $60^{\circ}$ with the line $\sqrt{3} x+y=1$ is

Options:
A $\sqrt{3} x+y-(2+3 \sqrt{3})=0$
B $\sqrt{3} x-y+(2-3 \sqrt{3})=0$
C $-\sqrt{3} x+y-(2-3 \sqrt{3})=0$
D $-\sqrt{3} x+y+(2-3 \sqrt{3})=0$

The correct answer is:
$\sqrt{3} x-y+(2-3 \sqrt{3})=0$

Solution:
$$
\begin{aligned}
& \sqrt{3} x+y=1 \\
& m_1=-\sqrt{3}
\end{aligned}
$$
Let slope of required line be $m_2$
$$
\begin{aligned}
& \tan 60^{\circ}=\left|\frac{m_1-m_2}{1+m_1 m_2}\right| \\
& \Rightarrow \sqrt{3}=\left|\frac{-\sqrt{3}-m_2}{1-\sqrt{3} m_2}\right|
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow \quad \sqrt{3} \cdot\left|1-\sqrt{3} m_2\right|=\left|-\sqrt{3}-m_2\right| \\
& \Rightarrow \quad 3\left(1-\sqrt{3} m_2\right)^2=\left(-\sqrt{3}-m_2\right)^2 \\
& \Rightarrow \quad 3\left(1+3 m_2^2-2 \sqrt{3} m_2\right)=3+m_2^2+2 \sqrt{3} m_2 \\
& \Rightarrow \quad 3+9 m_2^2-6 \sqrt{3} m_2=3+m_2^2+2 \sqrt{3} m_2 \\
& \Rightarrow \quad 8 m_2^2-8 \sqrt{3} m_2=0 \\
& \Rightarrow \quad m_2\left(8 m_2-8 \sqrt{3}\right)=0 \\
& \Rightarrow \quad m_2=0, \frac{8 \sqrt{3}}{8} \Rightarrow m_2=0, \sqrt{3}
\end{aligned}
$$
Taking $m_2=\sqrt{3}$
Equation of line
$$
\begin{aligned}
& y-2=\sqrt{3}(x-3) \\
\Rightarrow \quad & y-2=\sqrt{3} x-3 \sqrt{3} \\
\Rightarrow \quad & \sqrt{3} x-y+(2-3 \sqrt{3})=0 .
\end{aligned}
$$

Topic: Mathematics, Straight Lines, JEE Main
The equation of the straight line passing through the point $(3,2)$ and perpendicular to the line $y=x$ is

Options:
A $x-y=5$
B $x+y=5$
C $x+y=1$
D $x-y=1$

The correct answer is:
$x+y=5$

Solution:
Let the required equation is $y=-x+c$ which is perpendicular to $y=x$ and passes through $(3,2)$.
So $2=-3+c \Rightarrow c=5$. Hence required equation is $x+y=5$

Topic: Mathematics, Straight Lines, JEE Main
The equation of the straight line passing through the point $(4,3)$ and making intercepts on the co-ordinate axes whose sum is $-1$ is

Options:
A
$\frac{x}{2}+\frac{y}{3}=-1$ and $\frac{x}{-2}+\frac{y}{1}=-1$

B
$\frac{x}{2}-\frac{y}{3}=-1$ and $\frac{x}{-2}+\frac{y}{1}=-1$

C
$\frac{x}{2}+\frac{y}{3}=1$ and $\frac{x}{2}+\frac{y}{1}=1$

D
$\frac{x}{2}-\frac{y}{3}=1$ and $\frac{x}{-2}+\frac{y}{1}=1$

The correct answer is:
$\frac{x}{2}-\frac{y}{3}=1$ and $\frac{x}{-2}+\frac{y}{1}=1$

Solution:
$\frac{x}{a}+\frac{y}{b}=1$ where $a+b=-1$ and $\frac{4}{a}+\frac{3}{b}=1$
$\Rightarrow a=2, b=-3$ or $a=-2, b=1$
Hence $\frac{x}{2}-\frac{y}{3}=1$ and $\frac{x}{-2}+\frac{y}{1}=1$

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2004
The equation of the tangent at a point $P(t)$ where ' $t$ ' is any parameter to the parabola $y^2=4 a x$, is

Options:
A $y t=x+a t^2$
B $y=x t+a t^2$
C $y=x t+\frac{a}{t}$
D $y=t x$

The correct answer is:
$y t=x+a t^2$

Solution:
Any point on $y^2=4 a x$ is $\left(a t^2, 2 a t\right)$, then tangent is
\(y_1 . y=4 a(x+x 1) / 2\)
So, Tangent is 2aty= \(2 a\left(x+a t^{2}\right)\)
So, \(\Rightarrow \mathrm{yt}=\mathrm{x}+\mathrm{at}^{2}\).

Topic: Mathematics, Parabola, JEE Main
The equation of the tangent at the point $(1 / 4,1 / 4)$ of the ellipse $\frac{x^2}{4}+\frac{y^2}{12}=1$ is

Options:
A $3 x+y=48$
B $3 x+y=3$
C $3 x+y=16$
D None of these

The correct answer is:
$3 x+y=48$

Solution:
Equation of the tangent at point $(\frac{1}{4},\frac{1}{4})$ of ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{12}=1$
Is given as, $3 x+y=48$
The equation of a tangent can be given as $\frac{x \times y_1}{a}+\frac{y \times y_1}{b}=1$. Here,
$x 1=\frac{1}{4}$ and $y 1=\frac{1}{4}$ and the equation of the ellipse is given as $\frac{x^{2}}{4}+\frac{y^{2}}{12}=1$.
Putting the values in the above equation we get,
$\begin{aligned}
& \frac{\left(X \times \frac{1}{4}\right)}{4}+\frac{\left(y \times \frac{1}{4}\right)}{12}=1 \\
& \text {Or } \frac{X}{16}+\frac{y}{48}=1 \\
& \text {Or, } 48 x+16 y=16 \times 48
\end{aligned}$
Or, $3 x+y=48$ which is the required solution.
The equation of an ellipsis can also be calculated By parametric method.

Topic: Mathematics, Ellipse, JEE Main