Vector & 3D Geometry for Competitive Exam Preparation

Vector & 3D Geometry

3D Lines 3D Planes Scalar Product Vector Algebra Basics Vector Product

Q. If A = (1, 2, 3) and B = (4, 5, 6), what is the value of A · B?

A. 32
B. 30
C. 28
D. 26

Solution

A · B = 1*4 + 2*5 + 3*6 = 4 + 10 + 18 = 32.

Correct Answer: B — 30

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Q. If A = (1, 2, 3) and B = (k, k, k) are perpendicular, what is the value of k?

A. 1
B. 2
C. 3
D. 0

Solution

A · B = 1*k + 2*k + 3*k = 6k = 0. Thus, k = 0.

Correct Answer: D — 0

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Q. If A = (1, 2, 3) and B = (x, y, z) are such that A · B = 0, what is the condition for x, y, z?

A. x + 2y + 3z = 0
B. x - 2y + 3z = 0
C. x + 2y - 3z = 0
D. x - 2y - 3z = 0

Solution

A · B = 1*x + 2*y + 3*z = 0 gives the condition x + 2y + 3z = 0.

Correct Answer: A — x + 2y + 3z = 0

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Q. If A = (1, 2, 3) and B = (x, y, z) such that A · B = 14, find the value of x + y + z.

A. 5
B. 6
C. 7
D. 8

Solution

A · B = 1*x + 2*y + 3*z = 14. If we assume x = 2, y = 4, z = 2, then x + y + z = 8.

Correct Answer: C — 7

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Q. If A = (2, 0, -1) and B = (0, 3, 4), what is A · B?

A. -4
B. 0
C. 6
D. 8

Solution

A · B = 2*0 + 0*3 + (-1)*4 = 0 - 4 = -4.

Correct Answer: B — 0

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Q. If A = (2, 0, -1) and B = (0, 3, 4), what is the scalar product A · B?

A. -4
B. 0
C. 6
D. 8

Solution

A · B = 2*0 + 0*3 + (-1)*4 = 0 - 4 = -4.

Correct Answer: B — 0

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Q. If A = (2, 3) and B = (4, 5), what is the vector AB?

A. (2, 2)
B. (2, 3)
C. (4, 5)
D. (6, 8)

Solution

AB = B - A = (4 - 2, 5 - 3) = (2, 2)

Correct Answer: A — (2, 2)

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Q. If A = (2, 3) and B = (4, 7), find the vector AB.

A. (2, 4)
B. (2, 3)
C. (2, 1)
D. (2, 2)

Solution

Vector AB = B - A = (4 - 2, 7 - 3) = (2, 4).

Correct Answer: A — (2, 4)

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Q. If A = (2, 3) and B = (k, 1) are such that A · B = 10, find k.

A. 1
B. 2
C. 3
D. 4

Solution

A · B = 2k + 3*1 = 10. Thus, 2k + 3 = 10, leading to 2k = 7, k = 3.5.

Correct Answer: C — 3

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Q. If A = (2, 3, 4) and B = (0, 0, 0), what is A · B?

A. 0
B. 2
C. 3
D. 4

Solution

A · B = 2*0 + 3*0 + 4*0 = 0.

Correct Answer: A — 0

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Q. If A = (2, 3, 4) and B = (1, 0, -1), find the scalar product A · B.

A. -1
B. 0
C. 1
D. 10

Solution

A · B = 2*1 + 3*0 + 4*(-1) = 2 + 0 - 4 = -2.

Correct Answer: A — -1

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Q. If A = (2, 3, 4) and B = (1, 0, -1), find the vector product A × B.

A. (3, 6, -3)
B. (3, 4, -3)
C. (3, -4, 6)
D. (3, -6, 4)

Solution

A × B = |i  j  k|\n|2  3  4|\n|1  0 -1| = (3, 6, -3)

Correct Answer: A — (3, 6, -3)

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Q. If A = (2, 3, 4) and B = (1, 0, -1), what is the scalar product A · B?

A. -1
B. 0
C. 2
D. 10

Solution

A · B = 2*1 + 3*0 + 4*(-1) = 2 + 0 - 4 = -2.

Correct Answer: D — 10

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Q. If A = (2, 3, 4) and B = (k, 0, -1) are perpendicular, find k.

A. -4
B. -3
C. -2
D. -1

Solution

A · B = 2k + 3*0 + 4*(-1) = 0. Thus, 2k - 4 = 0, k = 2.

Correct Answer: A — -4

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Q. If A = (2, 3, 4) and B = (x, y, z) such that A · B = 20, find the value of x + y + z.

A. 5
B. 6
C. 7
D. 8

Solution

A · B = 2x + 3y + 4z = 20. If we assume x = 2, y = 2, z = 2, then 2*2 + 3*2 + 4*2 = 20, thus x + y + z = 6.

Correct Answer: C — 7

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Q. If A = (3, -1, 2) and B = (k, 4, -1) are orthogonal, find k.

A. -2
B. 1
C. 2
D. 3

Solution

A · B = 3k - 4 - 2 = 0. Thus, 3k - 6 = 0, k = 2.

Correct Answer: A — -2

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Q. If A = (3, -2, 1) and B = (4, 0, -1), what is the value of A · B?

A. -1
B. 0
C. 1
D. 10

Solution

A · B = 3*4 + (-2)*0 + 1*(-1) = 12 + 0 - 1 = 11.

Correct Answer: A — -1

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Q. If A = (a, b, c) and B = (1, 2, 3) such that A · B = 0, what is the relation between a, b, and c?

A. a + 2b + 3c = 0
B. a - 2b + 3c = 0
C. a + b + c = 0
D. a - b - c = 0

Solution

A · B = a*1 + b*2 + c*3 = 0. Thus, a + 2b + 3c = 0.

Correct Answer: A — a + 2b + 3c = 0

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Q. If A = (a, b, c) and B = (1, 2, 3), and A · B = 14, find a + b + c.

A. 5
B. 6
C. 7
D. 8

Solution

a*1 + b*2 + c*3 = 14. One possible solution is a = 2, b = 4, c = 0, so a + b + c = 6.

Correct Answer: C — 7

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Q. If A = (a, b, c) and B = (1, 2, 3), and A · B = 14, what is the equation?

A. a + 2b + 3c = 14
B. a - 2b + 3c = 14
C. a + 2b - 3c = 14
D. a - 2b - 3c = 14

Solution

A · B = a*1 + b*2 + c*3 = 14.

Correct Answer: A — a + 2b + 3c = 14

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Q. If A = (a, b, c) and B = (1, 2, 3), find the value of a if A · B = 10.

A. 1
B. 2
C. 3
D. 4

Solution

A · B = a*1 + b*2 + c*3 = 10. If b = 2 and c = 1, then a + 4 + 3 = 10, so a = 3.

Correct Answer: C — 3

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Q. If A = (a, b, c) and B = (1, 2, 3), what is the scalar product A · B?

A. a + 2b + 3c
B. a - 2b - 3c
C. a * b * c
D. a^2 + b^2 + c^2

Solution

A · B = a*1 + b*2 + c*3 = a + 2b + 3c.

Correct Answer: A — a + 2b + 3c

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Q. If A = (x, y) and B = (y, x), what is the scalar product A · B?

A. x^2 + y^2
B. xy
C. x^2 - y^2
D. 0

Solution

A · B = x*y + y*x = 2xy.

Correct Answer: A — x^2 + y^2

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Q. If A = (x, y, z) and B = (1, 1, 1) such that A · B = 6, find x + y + z.

A. 3
B. 6
C. 9
D. 12

Solution

A · B = x + y + z = 6. Thus, x + y + z = 6.

Correct Answer: B — 6

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Q. If A = (x, y, z) and B = (1, 1, 1), find the scalar product A · B.

A. x + y + z
B. x - y + z
C. x + y - z
D. x - y - z

Solution

A · B = x*1 + y*1 + z*1 = x + y + z.

Correct Answer: A — x + y + z

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Q. If A = (x, y, z) and B = (1, 2, 3), and A · B = 14, find the value of x + y + z.

A. 5
B. 6
C. 7
D. 8

Solution

A · B = x*1 + y*2 + z*3 = 14; Let x + y + z = k; We can find values satisfying this equation.

Correct Answer: C — 7

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Q. If A = (x, y, z) and B = (2, 2, 2), and A · B = 12, what is the value of x + y + z?

A. 6
B. 8
C. 10
D. 12

Solution

A · B = 2x + 2y + 2z = 12, thus x + y + z = 6.

Correct Answer: A — 6

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Q. If A = (x, y, z) and B = (2, 3, 4), and A · B = 10, find the value of x + y + z.

A. 1
B. 2
C. 3
D. 4

Solution

x*2 + y*3 + z*4 = 10. One possible solution is x = 1, y = 1, z = 1, so x + y + z = 3.

Correct Answer: C — 3

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Q. If A = (x, y, z) and B = (2, 3, 4), find the value of x if A · B = 10.

A. 1
B. 2
C. 3
D. 4

Solution

A · B = x*2 + y*3 + z*4 = 10. If y = 0 and z = 0, then x = 5.

Correct Answer: B — 2

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Q. If A = 1i + 1j + 1k and B = 2i + 2j + 2k, what is A · B?

A. 6
B. 4
C. 2
D. 0

Solution

A · B = (1)(2) + (1)(2) + (1)(2) = 2 + 2 + 2 = 6.

Correct Answer: A — 6

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Vector & 3D Geometry

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