datavorous/entrance-exam-dataset huggingface dataset:
If $\left|\begin{array}{lll}a & a^2 & 1+a^3 \\ b & b^2 & 1+b^3 \\ c & c^2 & 1+c^3\end{array}\right|=0$ and vectors $\left(1, a, a^2\right),\left(a, b, b^2\right)$ and $\left(a, c, c^2\right)$ are non-coplanar, then the product abc equals

Options:
A
0

B
2

C
$-1$

D
1

The correct answer is:
$-1$

Solution:
$\left|\begin{array}{lll}\mathrm{a} & \mathrm{a}^2 & 1+\mathrm{a}^3 \\ \mathrm{~b} & \mathrm{~b}^2 & 1+\mathrm{b}^3 \\ \mathrm{c} & \mathrm{c}^2 & 1+\mathrm{c}^3\end{array}\right|=0 \Rightarrow\left|\begin{array}{lll}\mathrm{a} & \mathrm{a}^2 & 1 \\ \mathrm{~b} & \mathrm{~b}^2 & 1 \\ \mathrm{c} & \mathrm{c}^2 & 1\end{array}\right|+\left|\begin{array}{ccc}\mathrm{a} & \mathrm{a}^2 & \mathrm{a}^3 \\ \mathrm{~b} & \mathrm{~b}^2 & \mathrm{~b}^3 \\ \mathrm{c} & \mathrm{c}^2 & \mathrm{c}^3\end{array}\right|=0$
$(a-b)(b-c)(c-a)+a b c(a-b)(b-c)(c-a)=0$
$(a b c+1)[(a-b)(b-c)(c-a)]=0$
As $\left|\begin{array}{lll}1 & \mathrm{a} & \mathrm{a}^2 \\ 1 & \mathrm{~b} & \mathrm{~b}^2 \\ 1 & \mathrm{c} & \mathrm{c}^2\end{array}\right| \neq 0$ (given condition) $\quad \therefore \mathrm{abc}=-1$

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2003
If $(\mathrm{a} \times \mathrm{b}) \times \overline{\mathrm{c}}=\overline{\mathrm{a}} \times(\overline{\mathrm{b}} \times \overline{\mathrm{c}})$, where $\overline{\mathrm{a}}, \overline{\mathrm{b}}$ and $\overline{\mathrm{c}}$ are any three vectors such that $\overline{\mathrm{a}} \cdot \overline{\mathrm{b}} \neq 0$, $\overline{\mathrm{b}} \cdot \overline{\mathrm{c}} \neq 0$, then $\mathrm{a}$ and $\mathrm{c}$ are

Options:
A
inclined at an angle of $\pi / 3$ between them

B
inclined at an angle of $\pi / 6$ between them

C
perpendicular

D
parallel

The correct answer is:
parallel

Solution:
$(\mathrm{a} \times \mathrm{b}) \times \overline{\mathrm{c}}=\overline{\mathrm{a}} \times(\overline{\mathrm{b}} \times \overline{\mathrm{c}}), \overline{\mathrm{a}} \cdot \overline{\mathrm{b}} \neq 0, \overline{\mathrm{b}} \cdot \overline{\mathrm{c}} \neq 0$
$\Rightarrow(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}) \overline{\mathrm{b}}-(\overline{\mathrm{b}} \cdot \overline{\mathrm{c}}) \overline{\mathrm{a}}=(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}) \overline{\mathrm{b}}-(\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}) \overline{\mathrm{c}}$
$(\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}) \overline{\mathrm{c}}=(\overline{\mathrm{b}} \cdot \overline{\mathrm{c}}) \overline{\mathrm{a}}$
$\overline{\mathrm{a}} \| \overline{\mathrm{c}}$

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2006
If $\alpha \neq \mathrm{a}, \beta \neq \mathrm{b}, \gamma \neq \mathrm{c}$ and $\left|\begin{array}{lll}\alpha & \mathrm{b} & \mathrm{c} \\ \mathrm{a} & \beta & \mathrm{c} \\ \mathrm{a} & \mathrm{b} & \gamma\end{array}\right|=0$, then $\frac{\mathrm{a}}{\alpha-\mathrm{a}}+\frac{\mathrm{b}}{\beta-\mathrm{b}}+\frac{\gamma}{\gamma-\mathrm{c}}$ is equal to:

Options:
A 3
B 0
C 1
D 2

The correct answer is:
0

Solution:
$\begin{aligned} & \mathrm{R}_1 \rightarrow \mathrm{R}_1-\mathrm{R}_2, \mathrm{R}_2 \rightarrow \mathrm{R}_2-\mathrm{R}_3 \\ & \left|\begin{array}{ccc}\alpha-\mathrm{a} & \mathrm{b}-\beta & 0 \\ 0 & \beta-\mathrm{b} & \mathrm{c}-\gamma \\ \mathrm{a} & \mathrm{b} & \gamma\end{array}\right|=0 \\ & (\alpha-\mathrm{a})(\gamma(\beta-\mathrm{b})-\mathrm{b}(\mathrm{c}-\gamma))-(\mathrm{b}-\beta)(-\mathrm{a}(\mathrm{c}-\gamma))=0 \\ & \gamma(\alpha-\mathrm{a})(\beta-\mathrm{b})-\mathrm{b}(\alpha-\mathrm{a})(\mathrm{c}-\gamma)+\mathrm{a}(\mathrm{b}-\beta)(\mathrm{c}-\gamma) \\ & \frac{\gamma}{\gamma-\mathrm{c}}+\frac{\mathrm{b}}{\beta-\mathrm{b}}+\frac{\mathrm{a}}{\alpha-\mathrm{a}}=0\end{aligned}$

Topic: Mathematics, Determinants, JEE Main, JEE Main 2024 (08 Apr Shift 2)
If $|\mathbf{a}|+|\mathbf{b}|=|\mathbf{c}|$ and $\mathbf{a}+\mathbf{b}=\mathbf{c}$, then the angle between $\mathbf{a}$ and $\mathbf{b}$ is

Options:
A $\frac{\pi}{2}$
B $\pi$
C 0
D None of these

The correct answer is:
0

Solution:
$\begin{aligned}
& \mathbf{a}+\mathbf{b}=\mathbf{c} \Rightarrow|\mathbf{a}|^2+|\mathbf{b}|^2+2 \mathbf{a} \cdot \mathbf{b}=|\mathbf{c}|^2 \\
& \text { and }|\mathbf{a}|+|\mathbf{b}|=|\mathbf{c}| \Rightarrow|\mathbf{a}|^2+|\mathbf{b}|^2+2|\mathbf{a}||\mathbf{b}|=|\mathbf{c}|^2 \\
& \therefore \mathbf{a} \cdot \mathbf{b}=|\mathbf{a}||\mathbf{b}| \Rightarrow \cos \theta=1 \Rightarrow \theta=0 .
\end{aligned}$

Topic: Mathematics, Vector Algebra, JEE Main
If $a, b, c$ and $d \in R$ such that $a^2+b^2=4$ and $c^2+d^2=2$ and if $(a+i b)^2=(c+i d)^2(x+i y)$, then $x^2+y^2$ is equal to

Options:
A $4$
B $3$
C $2$
D $1$

The correct answer is:
$4$

Solution:
Given, $(a+i b)^2=(c+i d)^2(x+i y)$
$\begin{array}{lc}\Rightarrow & |a+i b|^2=|c+i d|^2|x+i y| \\ \Rightarrow & a^2+b^2=\left(c^2+d^2\right)\left(\sqrt{x^2+y^2}\right) \\ \Rightarrow & 4=2\left(\sqrt{x^2+y^2}\right) \\ & \quad\left(\because a^2+b^2=4 \text { and } c^2+d^2=2 \text { given }\right) \\ \Rightarrow & \sqrt{x^2+y^2}=2 \\ \Rightarrow & x^2+y^2=4\end{array}$

Topic: Mathematics, Complex Number, JEE Main
If $A, B, C$ are any three arbitrary events, then which one of the following expressions shows that both $A$ and $B$ occur but not $C$ ?

Options:
A $A \cap B \cap \bar{C}$
B $A \cap B \cap \bar{C}$
C $A \cap B \cap C$
D $A \cap B \cap C$

The correct answer is:
$A \cap B \cap \bar{C}$

Solution:
If $\mathrm{A}, \mathrm{B}, \mathrm{C}$ are any three arbitary events then only expression $A \cap B \cap \bar{C}$ will show that both $A$ and $B$ occur but not $\mathrm{C}$.

Topic: Mathematics, Probability, JEE Main
If $\mathbf{a}, \mathbf{b}, \mathbf{c}$ are any three coplanar unit vectors, then

Options:
A $\mathbf{a} \cdot(\mathbf{b} \times \mathbf{c})=1$
B $\mathbf{a} .(\mathbf{b} \times \mathbf{c})=3$
C $(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c}=0$
D $(\mathbf{c} \times \mathbf{a}) \cdot \mathbf{b}=1$

The correct answer is:
$(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c}=0$

Solution:
$\quad \mathbf{a} \cdot(\mathbf{b} \times \mathbf{c})=0 \quad$ or $(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c}=0$.

Topic: Mathematics, Vector Algebra, JEE Main
If $\mathbf{a}, \mathbf{b}, \mathbf{c}$ are any vectors, then the true statement is

Options:
A $\mathbf{a} \times(\mathbf{b} \times \mathbf{c})=(\mathbf{a} \times \mathbf{b}) \times \mathbf{c}$
B $\mathbf{a} \times \mathbf{b}=\mathbf{b} \times \mathbf{a}$
C $\mathbf{a} \cdot(\mathbf{b} \times \mathbf{c})=\mathbf{a} \cdot \mathbf{b} \times \mathbf{a} \cdot \mathbf{c}$
D $\mathbf{a} \cdot(\mathbf{b}-\mathbf{c})=\mathbf{a} \cdot \mathbf{b}-\mathbf{a} \cdot \mathbf{c}$

The correct answer is:
$\mathbf{a} \cdot(\mathbf{b}-\mathbf{c})=\mathbf{a} \cdot \mathbf{b}-\mathbf{a} \cdot \mathbf{c}$

Solution:
The given statement is: $a .(b-c)=a . b-a . c$
To understand why this statement is true, let's break it down step by step.
1. Dot Product:
The dot product of two vectors $a$ and $b$ is denoted by a.b. It is a scalar quantity calculated by taking the sum of the products of the corresponding components of the vectors. The dot product is commutative, which means a.b = b.a.
2. Vector Subtraction:
The vector subtraction of two vectors $b$ and $c$ is denoted by $(b-c)$. It is performed by subtracting the corresponding components of the vectors. For example, if $b=(b_1, b_2, b_3)$ and $c=(c_1, c_2, c_3)$, then $(b-c)=(b_1-c_1, b_2-c_2, b_3-c_3)$.
3. Distributive Property:
The distributive property states that the dot product of a vector a with the vector sum ( $b+c$ ) is equal to the sum of the dot products of $a$ with $b$ and a with $c$. Mathematically, it can be represented as $a .(b+c)=a . b+a . c$.
Now, let's prove the given statement:
a. $(b-c)=a . b-a . c$
Using the distributive property, we can expand the left side of the equation:
a. $(b-c)=a \cdot b-a . c$
Now, let's expand the dot product of a with $(\mathrm{b}-\mathrm{c})$ :
$a \cdot(b-c)=a \cdot b-a \cdot c$
Using vector subtraction, we can expand (b - c):
$a \cdot(b-c)=a \cdot b-a \cdot c$
Now, let's distribute the dot product a.b and a.c:
$a \cdot(b-c)=a \cdot b-a \cdot c$
By substituting the values of a.b and a.c, we get:
$a \cdot(b-c)=a \cdot b-a \cdot c$
Therefore, the statement $a \cdot(b-c)=a \cdot b-a . c$ is true.

Topic: Mathematics, Vector Algebra, JEE Main
If $a, b, c$ are distinct $+v e$ real numbers and $a^2+b^2+c^2=1$ then $a b+b c+c a$ is

Options:
A
less than 1

B
equal to 1

C
greater than 1

D
any real no.

The correct answer is:
less than 1

Solution:
$a b+b c+c a=\frac{(a+b+c)^2-1}{2} < 1$

Topic: Mathematics, Basic of Mathematics, JEE Main, JEE Main 2002
If $a, b, c$ are in A.P. and $a^2, b^2, c^2$ are in G.P. such that $a < b < c$ and $a+b+c=\frac{3}{4}$, then the value of $a$ is

Options:
A
$\frac{1}{4}-\frac{1}{3 \sqrt{2}}$

B
$\frac{1}{4}-\frac{1}{4 \sqrt{2}}$

C
$\frac{1}{4}-\frac{1}{\sqrt{2}}$

D
$\frac{1}{4}-\frac{1}{2 \sqrt{2}}$

The correct answer is:
$\frac{1}{4}-\frac{1}{2 \sqrt{2}}$

Solution:
$\because a, b, c$ are in A.P. then
$$
a+c=2 b
$$
also it is given that,
$$
\begin{aligned}
&a+b+c=\frac{3}{4} \\
&\Rightarrow 2 b+b=\frac{3}{4} \Rightarrow b=\frac{1}{4}
\end{aligned}
$$
Again it is given that, $a^2, b^2, c^2$ are in G.P. then
$$
\left(b^2\right)^2=a^2 c^2 \Rightarrow a c=\pm \frac{1}{16}
$$
From (1), (2) and (3), we get;
$$
a \pm \frac{1}{16 a}=\frac{1}{2} \Rightarrow 16 a^2-8 a \pm 1=0
$$
Case I: $16 a^2-8 a+1=0$
$\Rightarrow a=\frac{1}{4}$ (not possible as $\left.a < b\right)$
Case II: $16 a^2-8 a-1=0 \Rightarrow a=\frac{8 \pm \sqrt{128}}{32}$
$$
\begin{aligned}
&\Rightarrow a=\frac{1}{4} \pm \frac{1}{2 \sqrt{2}} \\
&\therefore a=\frac{1}{4}-\frac{1}{2 \sqrt{2}} \quad(\because a < b)
\end{aligned}
$$

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
If $a, b, c$ are in A.P., then $\frac{(a-c)^2}{\left(b^2-a c\right)}=$

Options:
A $1$
B $2$
C $3$
D $4$

The correct answer is:
$4$

Solution:
If $a, b, c$ are in A.P. $\Rightarrow 2 b=a+c$
$\frac{(a-c)^2}{\left(b^2-a c\right)}$ $=\frac{(a-c)^2}{\left\{\left(\frac{a+c}{2}\right)^2-a c\right\}}$
so, $=\frac{(a-c)^2 4}{\left[a^2+c^2+2 a c-4 a c\right]}$ $=\frac{4(a-c)^2}{(a-c)^2}=4$
Trick : Put $a=1, b=2, c=3$, then the required value is $\frac{4}{1}=4$

Topic: Mathematics, Sequences and Series, JEE Main
If $a, b, c$ are in A.P., then $\frac{a}{b c}, \frac{1}{c}, \frac{2}{b}$ are in

Options:
A A.P.
B G.P.
C H.P.
D None of these

The correct answer is:
None of these

Solution:
$a, b, c$ be in A.P. then divide each term by $b c$, we get $\frac{a}{b c}, \frac{1}{c}, \frac{1}{b}$ be in A.P. But it is $\frac{2}{b}$ in place of $\frac{1}{b}$, so answer is (4)

Topic: Mathematics, Sequences and Series, JEE Main
If $a, b, c$ are in A.P., then the value of $\left|\begin{array}{ccc}x+2 & x+3 & x+a \\ x+4 & x+5 & x+b \\ x+6 & x+7 & x+c\end{array}\right|$ is

Options:
A $x-(a+b+c)$
B $9 x^2+a+b+c$
C $a+b+c$
D $0$

The correct answer is:
$0$

Solution:
$A=\left|\begin{array}{lll}x+2 & x+3 & x+a \\ x+4 & x+5 & x+b \\ x+6 & x+7 & x+c\end{array}\right|$
Applying $C_2 \rightarrow C_2-C_1$, we get,
$\Rightarrow A=\left|\begin{array}{lll}x+2 & 1 & x+a \\ x+4 & 1 & x+b \\ x+6 & 1 & x+c\end{array}\right|$
Applying $R_2 \rightarrow R_2-R_1$ and $R_3 \rightarrow R_3-R_1$
$\Rightarrow \quad A=\left|\begin{array}{ccc}x+2 & 1 & x+a \\ 2 & 0 & b-a \\ 4 & 0 & c-a\end{array}\right|=-1(2 c-2 a-4 b+4 a)$
$=2(2 b-c-a)$
$a, b, c$ are in A.P. $\Rightarrow A=0$.

Topic: Mathematics, Determinants, JEE Main
If A, B, C are in arithmetic progression and B=Ï4, then tanâ¡Atanâ¡Btanâ¡C=

Options:

The correct answer is:
1

Solution:
âµ Angles A, B, C are in arithmetic progression and â B=Ï4
then A=Ï4-Î¸ C=Ï4+Î¸
Hence tanâ¡Ï4-Î¸tanâ¡Ï4tanâ¡Ï4+Î¸
=1-tanâ¡Î¸1+tanâ¡Î¸.1.1+tanâ¡Î¸1-tanâ¡Î¸
=1

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
If a,b,c are in arithmetic progression and one root of the equation ax2+bx+c=0 is 2, then the other root is

Options:
A 34
B -34
C -54
D -52

The correct answer is:
-54

Solution:
Let Î± be the other root, given that 2 is a root. So,
aâ22+bâ2+c=0
â4a+2b+c=0
Also, a,b,c are in A.P.
So,
2b=a+c
â4a+a+c+c=0
â5a=-2c
âca=-52=2ÃÎ±
âÎ±=-54.

Topic: Mathematics, Sequences and Series, JEE Main
If $a, b, c$ are in G.P., $a-b, c-a, b-c$ are in H.P., then $a+4 b+c$ is equal to

Options:
A $0$
B $1$
C $-1$
D None of these

The correct answer is:
$0$

Solution:
As given
$b^2=a c$ and $\frac{2}{c-a}=\frac{1}{a-b}+\frac{1}{b-c}=\frac{a-c}{(a-b)(b-c)}$
$\Rightarrow \quad 2(a-b)(b-c)=-(a-c)^2$
$\Rightarrow \quad-(a-c)^2=2\left(a b-2 b^2+b c\right)=2 b\{a-2 \sqrt{a c}+c\}$
$\Rightarrow \quad-\{(\sqrt{a}-\sqrt{c})(\sqrt{a}+\sqrt{c})\}^2=2 b(\sqrt{a}-\sqrt{c})^2$
$\Rightarrow 2 b=-\{a+2 \sqrt{a c}+c\}=-a-2 b-c$ or $a+4 b+c=0$.

Topic: Mathematics, Sequences and Series, JEE Main
If $a, b, c$ are in G.P. and $x, y$ are the arithmetic means between $a, b$ and $b, c$ respectively, then $\frac{a}{x}+\frac{c}{y}$ is equal to

Options:
A $0$
B $1$
C $2$
D $\frac{1}{2}$

The correct answer is:
$2$

Solution:
Given that $a, b, c$ are in G.P.
So, $b^2=a c \ldots(i)$
$x=\frac{a+b}{2} \ldots(ii)$
$y=\frac{b+c}{2} \ldots(iii)$
Now $\frac{a}{x}+\frac{c}{y}=\frac{2 a}{a+b}+\frac{2 c}{b+c}=\frac{2(a b+b c+2 c a)}{a b+a c+b^2+b c}$
$=\frac{2(a b+b c+2 c a)}{(a b+a c+a c+b c)}=2,\left\{b^2=a c\right\}$
Trick : Let $a=1, b=2, c=4$, then obviously $x=\frac{3}{2}$ and $y=3$, then $\frac{1}{3 / 2}+\frac{4}{3}=2$

Topic: Mathematics, Sequences and Series, JEE Main
If $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are in $\mathrm{GP}$, then what is the value of
$\left|\begin{array}{lll}a & b & a+b \\ b & c & b+c \\ a+b & b+c & 0\end{array}\right|$

Options:
A 0
B 1
C $-1$
D None of these

The correct answer is:
0

Solution:
Since, $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are in $\mathrm{GP}$.
$\Rightarrow \mathrm{b}^{2}=\mathrm{ac}$
Explanding the determinant we get,
$\left|\begin{array}{ccc}a & b & a+b \\ b & c & b+c \\ a+b & b+c & 0\end{array}\right|$
$=a\left|\begin{array}{cc}c & b+c \\ b+c & 0\end{array}\right|-b\left|\begin{array}{cc}b & b+c \\ a+b & 0\end{array}\right|+(a+b)\left|\begin{array}{cc}b & c \\ a+b & b+c\end{array}\right|$
$=-a(b+c)^{2}+b(a+b)(b+c)+(a+b)\left(b^{2}+b c-a c-b c\right)$
$=-a\left(b^{2}+c^{2}+2 b c\right)+b\left(a b+a c+b^{2}+b c\right)$
$=-a b^{2}-a c^{2}-2 a b c+a b^{2}+2 a b c+b^{2} c \quad\left(\because b^{2}=a c\right)$
$=-a c^{2}+b^{2} c=-a c^{2}+a c \cdot c .$
$=-a c^{2}+a c^{2}=0$

Topic: Mathematics, Determinants, JEE Main
If $a, b, c$ are in H.P., then for all $n \in N$ the true statement is

Options:
A $a^n+c^n \lt 2 b^n$
B $a^n+c^n \gt 2 b^n$
C $a^n+c^n=2 b^n$
D None of the above

The correct answer is:
$a^n+c^n \gt 2 b^n$

Solution:
For two numbers $a$ and $c$
$\frac{a^n+c^n}{2} \gt \left(\frac{a+c}{2}\right)^n$ (Where $n \in N, n \gt 1$)
$A.M. \gt G.M. \gt H.M.$
$\therefore \quad \frac{a+b}{2} \gt b$ $a, b, c$ are in H.P.)
$\Rightarrow\left(\frac{a+c}{2}\right)^n \gt b^n \Rightarrow$ $\frac{a^n+c^n}{2} \gt \left(\frac{a+c}{2}\right)^n \gt b^n$

Topic: Mathematics, Sequences and Series, JEE Main
If $\mathbf{a}, \mathbf{b}, \mathbf{c}$ are mutually perpendicular unit vectors, then $|\mathbf{a}+\mathbf{b}+\mathbf{c}|=$

Options:
A $\sqrt{3}$
B 3
C 1
D 0

The correct answer is:
$\sqrt{3}$

Solution:
Three mutually perpendicular unit vectors $=\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$.
Therefore $|\mathbf{a}|=|\mathbf{b}|=|\mathbf{c}|=1$ and $\mathbf{a} \cdot \mathbf{b}=\mathbf{b} . \mathbf{c}=\mathbf{c} . \mathbf{a}=0$.
We know that
$\begin{aligned}
&|\mathbf{a}+\mathbf{b}+\mathbf{c}|^2=(\mathbf{a}+\mathbf{b}+\mathbf{c}) \cdot(\mathbf{a}+\mathbf{b}+\mathbf{c})=|\mathbf{a}|^2+|\mathbf{b}|^2 \\
&+|\mathbf{c}|^2+2(\mathbf{a} \cdot \mathbf{b}+\mathbf{b} \cdot \mathbf{c}+\mathbf{c} \cdot \mathbf{a})=1+1+1+0=3
\end{aligned}$
$\text { or }|\mathbf{a}+\mathbf{b}+\mathbf{c}|=\sqrt{3} \text {. }$

Topic: Mathematics, Vector Algebra, JEE Main
If a → , b → , c → are mutually perpendicular vectors having magnitude 1,2,3 respectively, then [ a → + b → + c → b → − a → c → ]=

Options:
A 0
B 6
C 12
D 18

The correct answer is:
12

Solution:
Given, |a→|=1, |b→|=2, |c→|=3 and
a→.b→=0=b→.c→=c→.a→ (as the three vectors are mutually perpendicular)

So,
[(a→+b→+c→)×(b→−a→)].c→
=[a→×b→−0+0−b→×a→+c→×b→-c→×→].c→
=[2(a→×b→).c→+(c→×b→).c→−(c→×a→).c→]
=2(a→×b→).c→+0−0
=2[a→ b→ c→]
= 2 · 1 · 2 · 3
= 1 2

Topic: Mathematics, Vector Algebra, JEE Main
If $\mathbf{a}, \mathbf{b}, \mathbf{c}$ are mutually perpendicular vectors of equal magnitudes, then the angle between the vectors $\mathbf{a}$ and $\mathbf{a}+\mathbf{b}+\mathbf{c}$ is

Options:
A $\frac{\pi}{3}$
B $\frac{\pi}{6}$
C $\cos ^{-1} \frac{1}{\sqrt{3}}$
D $\frac{\pi}{2}$

The correct answer is:
$\cos ^{-1} \frac{1}{\sqrt{3}}$

Solution:
Since ${ }^{\mathbf{a}, \mathbf{b}}$ and $\mathbf{c}$ are mutually perpendicular, so $\mathbf{a} \cdot \mathbf{b}=\mathbf{b} \cdot \mathbf{c}=\mathbf{c} \cdot \mathbf{a}=\mathbf{0}$
Angle between $\mathbf{a}$ and $\mathbf{a}+\mathbf{b}+\mathbf{c}$ is
$\cos \theta=\frac{\mathbf{a} \cdot(\mathbf{a}+\mathbf{b}+\mathbf{c})}{|\mathbf{a} \| \mathbf{a}+\mathbf{b}+\mathbf{c}|}$
Now $|\mathbf{a}|=|\mathbf{b}|=|\mathbf{c}|=\mathbf{a}$
$\begin{aligned}
& |\mathbf{a}+\mathbf{b}+\mathbf{c}|^2=\mathbf{a}^2+\mathbf{b}^2+\mathbf{c}^2+2 \mathbf{a} \cdot \mathbf{b}+2 \mathbf{b} \cdot \mathbf{c}+2 \mathbf{c} . \\
& \quad=a^2+a^2+a^2+0+0+0 \\
& \Rightarrow|\mathbf{a}+\mathbf{b}+\mathbf{c}|^2=3 a^2 \Rightarrow \mathbf{a}+\mathbf{b}+\mathbf{c} \mid=\sqrt{3} a
\end{aligned}$
Putting this value in (i), we get $\quad \theta=\cos ^{-1} \frac{1}{\sqrt{3}}$.

Topic: Mathematics, Vector Algebra, JEE Main
If aâ,bâ,câ are non coplanar non - zero vectors such that bâÃcâ=aâ, aâÃbâ=câ and câÃaâ=bâ, then which of the following is not true

Options:
A aâ=1
B aâ bâ câ=1

C aâ+bâ+câ=3
D aââ bââ câ

The correct answer is:
a→≠b→≠c→

Solution:
bâÃcâ=aâ â aâ bâ câ=aâ . aâ=aâ2
Similarly, aâ bâ câ=bâ2=câ2
â´ aâ=bâ=câ=aâ bâ câ
Now, bâÃ câ câÃ aâ aâÃbâ=aâ bâ câ2
â´ aâ bâ câ=aâ bâ câ2
But, aâ bâ cââ 0 (non-coplanar)
So aâ bâ câ=1
Hence aâ=bâ=câ=1

Topic: Mathematics, Vector Algebra, JEE Main
If $a, b, c$, are non zero complex numbers satisfying $a^2+b^2+c^2=0$ and
$\left|\begin{array}{ccc}b^2+c^2 & a b & a c \\ a b & c^2+a^2 & b c \\ a c & b c & a^2+b^2\end{array}\right|=k a^2 b^2 c^2$, then $k$ is equal to

Options:
A
1

B
3

C
4

D
2

The correct answer is:
4

Solution:
Let $\Delta=\left|\begin{array}{ccc}b^2+c^2 & a b & a c \\ a b & c^2+a^2 & b c \\ a c & b c & a^2+b^2\end{array}\right|$
Multiply $C_1$ by $a, C_2$ by b and $C_3$ by $\mathrm{c}$ and hence divide by $a b c$.
$$
=\frac{1}{a b c}\left|\begin{array}{ccc}
a\left(b^2+c^2\right. & \left.a b^2\right) & a c^2 \\
a^2 b & b\left(c^2+a^2\right. & \left.b c^2\right) \\
a^2 c & b^2 c & c\left(a^2+b^2\right.
\end{array}\right|
$$
Take out $a, b, c$ common from $R_1, R_2$ and $R_3$ respectively.
$$
\begin{aligned}
& \therefore \quad \Delta=\frac{a b c}{a b c}\left|\begin{array}{ccc}
b^2+c^2 & b^2 & c^2 \\
a^2 & c^2+a^2 & c^2 \\
a^2 & b^2 & a^2+b^2
\end{array}\right| \\
& \text { Apply } C_1 \rightarrow C_1-C_2-C_3 \\
& \Delta=\left|\begin{array}{ccc}
0 & b^2 & c^2 \\
-2 c^2 & c^2+a^2 & c^2 \\
-2 b^2 & b^2 & a^2+b^2
\end{array}\right| \\
& =-2\left|\begin{array}{ccc}
0 & b^2 & c^2 \\
c^2 & c^2+a^2 & c^2 \\
b^2 & b^2 & a^2+b^2
\end{array}\right| \\
&
\end{aligned}
$$
Apply $C_2-C_1$ and $C_3-C_1$
$$
\begin{aligned}
& =-2\left|\begin{array}{ccc}
0 & b^2 & c^2 \\
c^2 & a^2 & 0 \\
b^2 & 0 & a^2
\end{array}\right| \\
& =-2\left[-b^2\left(c^2 a^2\right)+c^2\left(-a^2 b^2\right)\right] \\
& =2 a^2 b^2 c^2+2 a^2 b^2 c^2=4 a^2 b^2 c^2 \\
& \text { But } \Delta=k a^2 b^2 c^2 \therefore k=4
\end{aligned}
$$

Topic: Mathematics, Determinants, JEE Main, JEE Main 2012 (19 May Online)
If a, b, c are non - zero real numbers and if the system of equations

a-1x=y+z

b-1y=x+z

c-1z=x+y

has a non - trivial solution, then ab+bc+ca equals :

Options:
A -1
B a+b+c
C abc
D 1

The correct answer is:
abc

Solution:
Given, a-1x=y+z⇒a-1x-y-z=0
b-1y=z+x⇒-x+b-1y-z=0
c-1z=y+x=-x-y+c-1z=0
For a non-trivial solution
a-1-1-1-1b-1-1-1-1c-1=0
⇒ 1-a1111-b1111-c=0
⇒ 1-a1-b1-c-1-11-c-1+1-1-b=0
⇒ 1-a+b+c+ab+bc+ca-abc-1+a+b+c=0
⇒ab+bc+ca=abc

Topic: Mathematics, Determinants, JEE Main, JEE Main 2014 (09 Apr Online)
If $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar vectors and $\lambda$ is a real number then $\left[\lambda(\vec{a}+\vec{b}) \lambda^2 \vec{b} \lambda \vec{c}\right]=[\vec{a} \vec{b}+\vec{c} \vec{b}]$ for

Options:
A
exactly one value of $\lambda$

B
no value of $\lambda$

C
exactly three values of $\lambda$

D
exactly two values of $\lambda$

The correct answer is:
no value of $\lambda$

Solution:
$$
\begin{aligned}
& {\left[\begin{array}{lll}
\lambda(\vec{a}+\vec{b}) & \lambda^2 \vec{b} & \lambda \vec{c}
\end{array}\right]=\left[\begin{array}{lll}
\vec{a} & \vec{b}+\vec{c} & \vec{b}
\end{array}\right]} \\
& \left|\begin{array}{ccc}
\lambda & \lambda & 0 \\
0 & \lambda^2 & 0 \\
0 & 0 & \lambda
\end{array}\right|=\left|\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 1 \\
0 & 1 & 0
\end{array}\right| \\
& \Rightarrow \lambda^4=-1 \\
&
\end{aligned}
$$
Hence no real value of $\lambda$.

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2005
If $\bar{a}, \bar{b}, \bar{c}$ are non-coplanar vectors and $\lambda$ is a real number, then the vectors $\overline{\mathrm{a}}+2 \overline{\mathrm{b}}+3 \overline{\mathrm{c}}, \lambda \overline{\mathrm{b}}+4 \overline{\mathrm{c}}$ and $(2 \lambda-1) \overline{\mathrm{c}}$ are non-coplanar for

Options:
A
all values of $\lambda$

B
all except one value of $\lambda$

C
all except two values of $\lambda$

D
no value of $\lambda$

The correct answer is:
all except two values of $\lambda$

Solution:
Condition for given three vectors to be coplanar is $\left|\begin{array}{ccc}1 & 2 & 3 \\ 0 & \lambda & 4 \\ 0 & 0 & 2 \lambda-1\end{array}\right|=0 \Rightarrow \lambda=0,1 / 2$
Hence given vectors will be non coplanar for all real values of $\lambda$ except $0,1 / 2$

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2004
If $\mathbf{a}, \mathbf{b}, \mathbf{c}$ are non-coplanar vectors and $\lambda$ is a real number, then the vectors $\mathbf{a}+\mathbf{2 b}+3 \mathbf{c}, \lambda \mathbf{b}+4 \mathbf{c}$ and $(2 \lambda-1) \mathbf{c}$ are non-coplanar for

Options:
A No value of $\lambda$
B All except one value of $\lambda$
C All except two values of $\lambda$
D All values of $\lambda$

The correct answer is:
All except two values of $\lambda$

Solution:
As $\mathbf{a}, \mathbf{b}, \mathbf{c}$ are non coplanar vectors
$\therefore[\mathbf{a} \mathbf{b} \mathbf{c}] \neq 0$
Now $\mathbf{a}+2 \mathbf{b}+3 \mathbf{c}, \lambda \mathbf{b}+4 \mathbf{c}$ and $(2 \lambda-1) \mathbf{c}$ will be non-coplanor, iff $(\mathbf{a}+2 \mathbf{b}+3 \mathbf{c}) .\{\lambda \mathbf{b}+4 \mathbf{c}) \times(2 \lambda-1) \mathbf{c}\} \neq 0$
i.e., $(\mathbf{a}+2 \mathbf{b}+3 \mathbf{c})\{\lambda(2 \lambda-1)(\mathbf{b} \times \mathbf{c})\} \neq 0$
i.e., $\lambda(2 \lambda-1)[\mathbf{a} \mathbf{b} \mathbf{c}] \neq 0, \ldots \quad \lambda \neq 0, \frac{1}{2}$
Thus given vectors will be non-coplanar for all values of $\lambda$ except two values, $\lambda=0$ and $\lambda=1 / 2$.
Trick : For coplanarity,
$\left|\begin{array}{ccc}
1 & 2 & 3 \\
0 & \lambda & 4 \\
0 & 0 & 2 \lambda-1
\end{array}\right|=0 \Rightarrow \lambda=0, \frac{1}{2}$
$\therefore$ All values except two values of $\lambda=0, \frac{1}{2}$.

Topic: Mathematics, Vector Algebra, JEE Main