Let a ? = 2 i ^ + j ^ – k ^ , b ? = ( ( a ? × ( i ^ + j ^ ) ) × i ^ ) × i ^ . Then the square of the projection of a ? on b ? is : [2024]

Question

Let $\vec{a} = 2 \hat{i} + \hat{j} - \hat{k}$ , $\vec{b} = ((\vec{a} \times (\hat{i} + \hat{j})) \times \hat{i}) \times \hat{i}$ . Then the square of the projection of $\vec{a}$ on $\vec{b}$ is : [2024]

Smart Achievers · Accepted Answer

(4)

We have, $\vec{a} = 2 \hat{i} + \hat{j} - \hat{k}$

and $\vec{b} = ((\vec{a} \times (\hat{i} + \hat{j})) \times \hat{i}) \times \hat{i}$

$=$ $(| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & - 1 \\ 1 & 1 & 0 \end{matrix} | \times \hat{i})$ $\times \hat{i} = ((\hat{i} - \hat{j} + \hat{k}) \times \hat{i})$

$=$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & - 1 & 1 \\ 1 & 0 & 0 \end{matrix} |$ $\times \hat{i} = (\hat{j} + \hat{k}) \times \hat{i} = - \hat{k} + \hat{j}$

$\Rightarrow \vec{b} = \hat{j} - \hat{k}$

So, projection of $\vec{a}$ on $\vec{b}$ = $\frac{\vec{a} \cdot \vec{b}}{| \vec{b} |} = \frac{2}{\sqrt{2}} = \sqrt{2}$

$∴$ Square of projection = 2.

Solution

Q.
Let $\vec{a} = 2 \hat{i} + \hat{j} - \hat{k}$ , $\vec{b} = ((\vec{a} \times (\hat{i} + \hat{j})) \times \hat{i}) \times \hat{i}$ . Then the square of the projection of $\vec{a}$ on $\vec{b}$ is : [2024]

1 $\frac{1}{3}$

2 $\frac{1}{5}$

3 $\frac{2}{3}$

4 2

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Solution

Q. Let a→=2i^+j^–k^, b→=((a→×(i^+j^))×i^)×i^. Then the square of the projection of a→ on b→ is : [2024]

1 13

2 15

3 23

4 2

Ans. (4) We have, a→=2i^+j^–k^ and b→=((a→×(i^+j^))×i^)×i^ =(|i^j^k^21–1110|×i^)×i^=((i^–j^+k^)×i^) =|i^j^k^1–11100|×i^=(j^+k^)×i^=–k^+j^ ⇒ b→=j^–k^ So, projection of a→ on b→ = a→·b→|b→|=22=2 ∴ Square of projection = 2.

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Q.
Let $\vec{a} = 2 \hat{i} + \hat{j} - \hat{k}$ , $\vec{b} = ((\vec{a} \times (\hat{i} + \hat{j})) \times \hat{i}) \times \hat{i}$ . Then the square of the projection of $\vec{a}$ on $\vec{b}$ is : [2024]

1 $\frac{1}{3}$

2 $\frac{1}{5}$

3 $\frac{2}{3}$