If is a non-zero vector such its projections on the vectors , and are equal, then a unit vector along is: [2025]

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Solution

Q.
If $\vec{a}$ is a non-zero vector such its projections on the vectors $2 \hat{i} - \hat{j} + 2 \hat{k}$ , $\hat{i} + 2 \hat{j} - 2 \hat{k}$ and $\hat{k}$ are equal, then a unit vector along $\vec{a}$ is: [2025]

1 $\frac{1}{\sqrt{155}} (7 \hat{i} + 9 \hat{j} + 5 \hat{k})$

2 $\frac{1}{\sqrt{155}} (7 \hat{i} + 9 \hat{j} - 5 \hat{k})$

3 $\frac{1}{\sqrt{155}} (- 7 \hat{i} + 9 \hat{j} - 5 \hat{k})$

4 $\frac{1}{\sqrt{155}} (- 7 \hat{i} + 9 \hat{j} + 5 \hat{k})$

Ans.
(1)

Let $\vec{a} = x \hat{i} + y \hat{j} + z \hat{k}, \vec{u} = 2 \hat{i} - \hat{j} + 2 \hat{k}, \vec{v} = \hat{i} + 2 \hat{j} - 2 \hat{k}$ and $\vec{w} = \hat{k}$

When, $\frac{\vec{a} \cdot \vec{u}}{| \vec{u} |} = \frac{\vec{a} \cdot \vec{v}}{| \vec{v} |} = \frac{\vec{a} \cdot \vec{w}}{| \vec{w} |}$

Now, $\frac{\vec{a} \cdot \vec{u}}{| \vec{u} |} = \frac{\vec{a} \cdot \vec{w}}{| \vec{w} |}$

$\Rightarrow \frac{(x \hat{i} + y \hat{j} + z \hat{k}) \cdot (2 \hat{i} - \hat{j} + 2 \hat{k})}{3} = \frac{(x \hat{i} + y \hat{j} + z \hat{k}) \cdot (\hat{k})}{1}$

$\Rightarrow 2 x - y + 2 z = 3 z$

$\Rightarrow 2 x - y - z = 0$ ... (i)

Also, $\frac{\vec{a} \cdot \vec{v}}{| \vec{v} |} = \frac{\vec{a} \cdot \vec{w}}{| \vec{w} |}$

$\Rightarrow \frac{x + 2 y - 2 z}{3} = z$

$\Rightarrow x + 2 y - 5 z = 0$ ... (ii)

and $\frac{\vec{a} \cdot \vec{u}}{| \vec{u} |} = \frac{\vec{a} \cdot \vec{v}}{| \vec{v} |}$

$\Rightarrow 2 x - y + 2 z = x + 2 y - 2 z$

$\Rightarrow x - 3 y + 4 z = 0$ ... (iii)

From (i), (ii) and (iii), we get x = 7, y = 9 and z = 5

$\vec{a} = \frac{7 \hat{i} + 9 \hat{j} + 5 \hat{k}}{\sqrt{{(7)}^{2} + {(9)}^{2} + {(5)}^{2}}} = \frac{7 \hat{i} + 9 \hat{j} + 5 \hat{k}}{\sqrt{155}}$

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