Let $\vec{b} = 4 \hat{i} + 3 \hat{j}$ and $\vec{c}$ be a vector perpendicular to
$\vec{b}$ and lying in the xy-plane. A vector in the xy-plane having
projections 1 and 2 along $\vec{b}$ and $\vec{c}$ is

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$\hat{i} - 2 \hat{j}$

none of these

$2 \hat{i} + 11 \hat{j}$

$2 \hat{i} - \hat{j}$

answer is A.

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Detailed Solution

Let the required vector be $\vec{a} = x \hat{i} + y \hat{j}$ . It is given

that the vector $\vec{c}$ is in xy-plane and is perpendicular to

$\vec{b} = 4 \hat{i} + 3 \hat{j} .$ Therefore,

by $\vec{c}$ and $\vec{d}$

$\vec{c} = λ (3 \hat{i} - 4 \hat{j}) \Rightarrow \hat{c} = \frac{1}{5} (3 \hat{i} - 4 \hat{j})$

Now, $\vec{a} . \hat{b} = 1$ and $\vec{a} \cdot \hat{c} = 2$

$\Rightarrow 4 x + 3 y = 5$ and $3 x - 4 y = 10$

Solving these two equations, we get $x = 2, y = - 1$

Hence, $\vec{b} = 2 \hat{i} - \hat{j}$

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