Question

Let $\mathbf{a} = 4\mathbf{i} + 3\mathbf{j}$ and $\mathbf{b}$ be two perpendicular vectors in the XOY-plane. A vector $\mathbf{c}$ in the same plane and having projections 1 and 2 respectively on $\mathbf{a}$ and $\mathbf{b}$ is

• A. $\mathbf{i} + 2\mathbf{j}$
• B. $2\mathbf{i} + \mathbf{j}$
• C. $\mathbf{i} - 2\mathbf{j}$
• D. $2\mathbf{i} - \mathbf{j}$

Hint:

For vector projections, use $\frac{\mathbf{c} \cdot \mathbf{a}}{|\mathbf{a}|}$. Solve the resulting system of equations from perpendicularity and projection conditions.

Answer 1

The Correct Option is D

Given $\mathbf{a} = 4\mathbf{i} + 3\mathbf{j}$, let $\mathbf{b} = x\mathbf{i} + y\mathbf{j}$ be perpendicular to $\mathbf{a}$: \[ \mathbf{a} \cdot \mathbf{b} = 4x + 3y = 0 \implies y = -\frac{4}{3}x \] Choose $x = 3$, then $y = -4$, so $\mathbf{b} = 3\mathbf{i} - 4\mathbf{j}$. Magnitude of $\mathbf{a}$: \[ |\mathbf{a}| = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = 5 \] Magnitude of $\mathbf{b}$: \[ |\mathbf{b}| = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = 5 \] Let $\mathbf{c} = \alpha \mathbf{i} + \beta \mathbf{j}$. The projection of $\mathbf{c}$ on $\mathbf{a}$ is 1: \[ \frac{\mathbf{c} \cdot \mathbf{a}}{|\mathbf{a}|} = \frac{4\alpha + 3\beta}{5} = 1 \implies 4\alpha + 3\beta = 5 \] The projection of $\mathbf{c}$ on $\mathbf{b}$ is 2: \[ \frac{\mathbf{c} \cdot \mathbf{b}}{|\mathbf{b}|} = \frac{3\alpha - 4\beta}{5} = 2 \implies 3\alpha - 4\beta = 10 \] Solve the system: \[ 4\alpha + 3\beta = 5 \quad (1) \] \[ 3\alpha - 4\beta = 10 \quad (2) \] Multiply (1) by 4 and (2) by 3: \[ 16\alpha + 12\beta = 20 \] \[ 9\alpha - 12\beta = 30 \] Add: \[ 25\alpha = 50 \implies \alpha = 2 \] Substitute into (1): \[ 4 \cdot 2 + 3\beta = 5 \implies 8 + 3\beta = 5 \implies 3\beta = -3 \implies \beta = -1 \] Thus, $\mathbf{c} = 2\mathbf{i} - \mathbf{j}$. Option (4) is correct. Options (1), (2), and (3) do not satisfy the projection conditions.