Mathematics

Problem

Match the items of Column I with items of Column II.

Column I	Column II
A. Given two vectors $\mathbf{a} = 2\hat{i} - 3\hat{j} + 6\hat{k}$ , $\mathbf{b} = -2\hat{i} + 2\hat{j} - \hat{k}$ and $\lambda = \frac{\text{Projection of } \mathbf{a} \text{ on } \mathbf{b}}{\text{Projection of } \mathbf{b} \text{ on } \mathbf{a}}$ , then the value of $3\lambda$ is	p. 0
B. If $\mathbf{a} = \hat{i} + 2\hat{j} + 3\hat{k}$ , $\mathbf{b} = -\hat{i} + 2\hat{j} + \hat{k}$ , $\mathbf{c} = 3\hat{i} + \hat{j}$ and $\mathbf{a} + p\mathbf{b}$ is normal to $\mathbf{c}$ , then $p$ is	q. 7
C. Let $\mathbf{a}$ , $\mathbf{b}$ , $\mathbf{c}$ be three non-zero vectors such that $\mathbf{a} + \mathbf{b} + \mathbf{c} = 0$ , then $\lambda(\mathbf{a} \times \mathbf{b}) + \mathbf{c} \times \mathbf{b} + \mathbf{a} \times \mathbf{c} = 0$ , where $\lambda$ is equal to	r. 5
D. The points whose position vectors are $p\hat{i} + q\hat{j} + r\hat{k}$ , $q\hat{i} + r\hat{j} + p\hat{k}$ and $r\hat{i} + p\hat{j} + q\hat{k}$ are collinear, then the value of $(p^2 + q^2 + r^2 - pq - qr - rp)$ is	s. 2

Intuitive explanation

Vector projections measure how much one vector "points in the direction" of another—use the dot product formula $\text{proj}_{\mathbf{b}}\mathbf{a} = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{b}|}$ to find their ratio. For perpendicularity, remember that two vectors are orthogonal when their dot product equals zero. The vector triple product identity and the constraint $\mathbf{a} + \mathbf{b} + \mathbf{c} = 0$ let you simplify cross product sums. Collinearity means the vectors connecting three points are parallel, which constrains the relationship between $p$ , $q$ , and $r$ .

Intuitive explanation

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