23. Proyeksi Vektor:
Given:

Vector p = (-1, 2, 4)
Vector q = (0, -2, 2)
Problem:
Find the projection of (p + q) onto q.

Solution:

Calculate p + q:
[
\mathbf{p} + \mathbf{q} = (-1 + 0, 2 + (-2), 4 + 2) = (-1, 0, 6)
]
Projection Formula:
The projection of vector a onto vector b is given by:
[
\text{Proj}_{\mathbf{b}} \mathbf{a} = \left( \frac{\mathbf{a} \cdot \mathbf{b}}{\mathbf{b} \cdot \mathbf{b}} \right) \mathbf{b}
]
Compute Dot Products:
[
\mathbf{a} \cdot \mathbf{b} = (-1)(0) + (0)(-2) + (6)(2) = 0 + 0 + 12 = 12
]
[
\mathbf{b} \cdot \mathbf{b} = 0^2 + (-2)^2 + 2^2 = 0 + 4 + 4 = 8
]
Calculate Projection:
[
\text{Proj}_{\mathbf{q}} (\mathbf{p} + \mathbf{q}) = \left( \frac{12}{8} \right) \mathbf{q} = \frac{3}{2} \times (0, -2, 2) = (0, -3, 3)
]
Expressed in unit vectors:
[
0\mathbf{i} - 3\mathbf{j} + 3\mathbf{k} = -3\mathbf{j} + 3\mathbf{k} = 3(-\mathbf{j} + \mathbf{k})
]
Interpreting the Options:
The calculated projection is -3j + 3k, which is a scalar multiple of option A. -j + k. Typically, projection directions are considered up to a scalar multiple, especially when only direction is of interest.
Answer:
A. -j + k

24. Panjang Proyeksi Vektor:
Given:

Vector a = (x – 3, -2, 4)
Vector b = (-2, 1, 1)
Length of projection of a on b is √6.
Problem:
Find the value of x that satisfies the above condition.

Solution:

Projection Length Formula:
The length of the projection of a onto b is:
[
\left| \text{Proj}_{\mathbf{b}} \mathbf{a} \right| = \frac{|\mathbf{a} \cdot \mathbf{b}|}{|\mathbf{b}|}
]
Given that this length is √6:
[
\frac{|\mathbf{a} \cdot \mathbf{b}|}{|\mathbf{b}|} = \sqrt{6}
]
Compute Dot Product a·b:
[
\mathbf{a} \cdot \mathbf{b} = (x - 3)(-2) + (-2)(1) + (4)(1) = -2x + 6 - 2 + 4 = -2x + 8
]
Compute Magnitude of b:
[
|\mathbf{b}| = \sqrt{(-2)^2 + 1^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6}
]
Set Up the Equation:
[
\frac{|-2x + 8|}{\sqrt{6}} = \sqrt{6} \implies |-2x + 8| = 6
]
This leads to two cases:
Case 1: (-2x + 8 = 6 \implies -2x = -2 \implies x = 1) (Not among the options)
Case 2: (-2x + 8 = -6 \implies -2x = -14 \implies x = 7)
Answer:
C. 7

25. Polinomial dan Vektor Tegak Lurus:
Given:

Vector a = (2x + 1, x - 3, 4)
Vector b = (x² + 1, 3x, 3x + 2)
Problem:
Find the value of ( x_1(x_2 - 2x_3) + x_2(x_3 - 2x_1) + x_3(x_1 - 2x_2) ) where ( x_1, x_2, x_3 ) are the roots of the polynomial formed when a and b are orthogonal.

Solution:

Orthogonality Condition:
Vectors a and b are orthogonal if their dot product is zero:
[
\mathbf{a} \cdot \mathbf{b} = 0
]
Compute Dot Product:
[
(2x + 1)(x^2 + 1) + (x - 3)(3x) + 4(3x + 2) = 0
]
Expand each term:
[
(2x + 1)(x^2 + 1) = 2x^3 + 2x + x^2 + 1
]
[
(x - 3)(3x) = 3x^2 - 9x
]
[
4(3x + 2) = 12x + 8
]
Combine all terms:
[
2x^3 + x^2 + 2x + 1 + 3x^2 - 9x + 12x + 8 = 2x^3 + 4x^2 + 5x + 9 = 0
]
The polynomial is:
[
2x^3 + 4x^2 + 5x + 9 = 0
]
Apply Vieta's Formulas:
For a cubic equation ( ax^3 + bx^2 + cx + d = 0 ):
Sum of roots: ( x_1 + x_2 + x_3 = -\frac{b}{a} = -\frac{4}{2} = -2 )
Sum of product of roots two at a time: ( x_1x_2 + x_1x_3 + x_2x_3 = \frac{c}{a} = \frac{5}{2} )
Product of roots: ( x_1x_2x_3 = -\frac{d}{a} = -\frac{9}{2} )
Simplify the Given Expression:
[
x_1(x_2 - 2x_3) + x_2(x_3 - 2x_1) + x_3(x_1 - 2x_2)
]
Expand the terms:
[
= x_1x_2 - 2x_1x_3 + x_2x_3 - 2x_1x_2 + x_3x_1 - 2x_2x_3
]
Combine like terms:
[
= (-x_1x_2) + (-x_1x_3) + (-x_2x_3) = -(x_1x_2 + x_1x_3 + x_2x_3)
]
Substitute the sum of the product of roots:
[
= -\left( \frac{5}{2} \right) = -\frac{5}{2}
]
Answer:
A. -5/2