Class 12 Vector Algebra Mock Test (CBSE)

Topics in Vector Algebra

• Vectors and scalars
• Types of vectors
• Addition of vectors
• Multiplication of a vector by a scalar
• Components of a vector
• Section formula
• Scalar (dot) product
• Projection of a vector
• Vector (cross) product

Vector Algebra practice questions with answers

1. Q1Assertion and reason·1 mark
   Assertion (A): Equal vectors always have the same magnitude and direction, regardless of the positions of their initial points. Reason (R): Two vectors are said to be equal if they have the same magnitude and direction, even if they have different initial points.

   1. (a)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
   2. (b)Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
   3. (c)Assertion (A) is true but Reason (R) is false.
   4. (d)Assertion (A) is false but Reason (R) is true.
   Show answer and step marks

   1. Correct option: (a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).+1

   Answer: (a)
2. Q2Multiple choice·1 mark
   Which of the following physical quantities is a vector quantity?

   1. (a)Speed
   2. (b)Distance
   3. (c)Time
   4. (d)Force
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   1. Correct option: (d) Force+1

   Answer: (d)
3. Q3Multiple choice·1 mark
   If $\vec{a}$ and $\vec{b}$ are two vectors such that $|\vec{a}| = 2$, $|\vec{b}| = 1$, and $\vec{a} \cdot \vec{b} = 1$, then the value of $|\vec{a} \times \vec{b}|$ is:

   1. (a)1
   2. (b)$2\sqrt{3}$
   3. (c)2
   4. (d)$\sqrt{3}$
   Show answer and step marks

   1. Correct option: (d) $\sqrt{3}$+1

   Answer: (d)
4. Q4Short answer·3 marks
   If $\vec{a} = 3\hat{i} - 2\hat{j} + \hat{k}$, $\vec{b} = \hat{i} - 3\hat{j} + 5\hat{k}$, and $\vec{c} = 2\hat{i} + \hat{j} - 4\hat{k}$, show that $\vec{b} + \vec{c} = \vec{a}$ and hence prove that the vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ form the sides of a right-angled triangle.
   Show answer and step marks

   1. Compute $\vec{b} + \vec{c} = (\hat{i} - 3\hat{j} + 5\hat{k}) + (2\hat{i} + \hat{j} - 4\hat{k}) = 3\hat{i} - 2\hat{j} + \hat{k} = \vec{a}$+1
   2. Calculate the magnitudes: $|\vec{a}| = \sqrt{3^2 + (-2)^2 + 1^2} = \sqrt{14}$, $|\vec{b}| = \sqrt{1^2 + (-3)^2 + 5^2} = \sqrt{35}$, and $|\vec{c}| = \sqrt{2^2 + 1^2 + (-4)^2} = \sqrt{21}$+1
   3. Check the relation $|\vec{b}|^2 = |\vec{a}|^2 + |\vec{c}|^2$ ($35 = 14 + 21$), which holds true, hence the vectors form a right-angled triangle (equivalently, $\vec{a} \cdot \vec{c} = 6 - 2 - 4 = 0$, so the right angle is between $\vec{a}$ and $\vec{c}$).+1

   Answer: $\vec{b} + \vec{c} = 3\hat{i} - 2\hat{j} + \hat{k} = \vec{a}$, so the vectors form a triangle; $|\vec{b}|^2 = 35 = 14 + 21 = |\vec{a}|^2 + |\vec{c}|^2$ (or $\vec{a} \cdot \vec{c} = 0$), so it is right-angled.

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