Let a = 4 i ^ + 3 j ^ and b = 3 i ^ - 4 j ^ + 5 k ^ . If c is a vector such that c · ( a × b ) + 25 = 0 , c · ( i ^ + j ^ + k ^ ) = 4 and projection of c on a is 1, then the projection of c on b equals: [2023]

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Solution

Q.
Let $\vec{a} = 4 \hat{i} + 3 \hat{j}$ and $\vec{b} = 3 \hat{i} - 4 \hat{j} + 5 \hat{k} .$ If $\vec{c}$ is a vector such that $\vec{c} \cdot (\vec{a} \times \vec{b}) + 25 = 0,$ $\vec{c} \cdot (\hat{i} + \hat{j} + \hat{k}) = 4$ and projection of $\vec{c}$ on $\vec{a}$ is 1, then the projection of $\vec{c}$ on $\vec{b}$ equals: [2023]

1 $\frac{1}{\sqrt{2}}$

2 $\frac{1}{5}$

3 $\frac{3}{\sqrt{2}}$

4 $\frac{5}{\sqrt{2}}$

Ans.
(4)

We have $\vec{a} = 4 \hat{i} + 3 \hat{j} + 0 \cdot \hat{k}$

and $\vec{b} = 3 \hat{i} - 4 \hat{j} + 5 \hat{k}$

such that $\vec{c} \cdot (\vec{a} \times \vec{b}) + 25 = 0$ ...(i)

$\vec{c} \cdot (\hat{i} + \hat{j} + \hat{k}) = 4$

We must have $\vec{a} \times \vec{b} = - 15 \hat{i} - 20 \hat{j} - 25 \hat{k}$

Let $\vec{c} = x \hat{i} + y \hat{j} + z \hat{k}$

$\Rightarrow 15 x - 20 y - 25 z + 25 = 0 (from eq. (i))$

$\Rightarrow 3 x - 4 y - 5 z = - 5$

Also, $x + y + z = 4$

and $\frac{\vec{c} \cdot \vec{a}}{| \vec{a} |} = 1$ $\Rightarrow 4 x + 3 y = 5$ $\Rightarrow \vec{c} = 2 \hat{i} - \hat{j} + 3 \hat{k}$

$∴$ Projection of $\vec{c}$ on $\vec{b}$ $= \frac{\vec{c} \cdot \vec{b}}{| \vec{b} |} = \frac{25}{5 \sqrt{2}} = \frac{5}{\sqrt{2}}$

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