Chapter 10: Vector Algebra — Online MCQ Test
MATHS · CLASS 12 INTER II YEAR · Andhra State Board
Practice Chapter 10: Vector Algebra with a free chapter-wise online MCQ test.
This chapter covers: Vector Definition: A vector is a mathematical quantity that possesses both a magnitude (length) and a definite direction in space.Direction Cosines: The direction cosines (l, m, n)....
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Chapter 10: Vector Algebra — Important Questions & Answers
A vector is a mathematical quantity that possesses which of the following characteristics?
- A. Only magnitude
- B. Only direction
- C. Both magnitude and direction
- D. Neither magnitude nor direction
Answer: C. Both magnitude and direction
By definition, a vector has both magnitude (length) and a definite direction in space.
By definition, a vector has both magnitude (length) and a definite direction in space.
Find the magnitude of the vector vec{a} = 3î + 4ĵ.
- A. 5
- B. 7
- C. 12
- D. 25
Answer: A. 5
Magnitude |vec{a}| = √(3² + 4²) = √(9 + 16) = √25 = 5.
Magnitude |vec{a}| = √(3² + 4²) = √(9 + 16) = √25 = 5.
Calculate vec{a} · vec{b} for vec{a} = 2î + 3ĵ and vec{b} = 4î + 5ĵ.
- A. 8
- B. 15
- C. 23
- D. 35
Answer: C. 23
vec{a} · vec{b} = (2)(4) + (3)(5) = 8 + 15 = 23.
vec{a} · vec{b} = (2)(4) + (3)(5) = 8 + 15 = 23.
Three points A, B, and C are collinear if which condition is satisfied?
- A. vec{AB} · vec{BC} = 0
- B. vec{AB} × vec{BC} = 0
- C. |vec{AB}| = |vec{BC}|
- D. vec{AB} + vec{BC} = vec{AC}
Answer: B. vec{AB} × vec{BC} = 0
Three points are collinear if the vectors formed by them are parallel, which means their cross product is zero.
Three points are collinear if the vectors formed by them are parallel, which means their cross product is zero.
The vectors vec{a} = xî + ĵ + k̂, vec{b} = î + yĵ + k̂, and vec{c} = î + ĵ + zk̂ are mutually perpendicular. Which statement is correct?
- A. x = y = z = -1
- B. x + y + z = -2
- C. xyz = 1
- D. x = 1, y = 1, z = 1
Answer: B. x + y + z = -2
From vec{a}·vec{b}=0, vec{b}·vec{c}=0, vec{c}·vec{a}=0, we get x+y = -1, y+z = -1, z+x = -1. Solving: x+y+z = -2.
From vec{a}·vec{b}=0, vec{b}·vec{c}=0, vec{c}·vec{a}=0, we get x+y = -1, y+z = -1, z+x = -1. Solving: x+y+z = -2.