Let a = 3 i – j + 2 k , b = a × ( i – 2 k ) and c = b × k . Then the projection of c – 2 j on a is : [2025]

Question

Let $\vec{a} = 3 \hat{i} - \hat{j} + 2 \hat{k}$ , $\vec{b} = \vec{a} \times (\hat{i} - 2 \hat{k})$ and $\vec{c} = \vec{b} \times \hat{k}$ . Then the projection of $\vec{c} - 2 \hat{j}$ on $\vec{a}$ is : [2025]

Smart Achievers · Accepted Answer

(3)

Given: $\vec{a} = 3 \hat{i} - \hat{j} + 2 \hat{k}$

and $\vec{b} = \vec{a} \times (\hat{i} - 2 \hat{k}) =$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & - 1 & 2 \\ 1 & 0 & - 2 \end{matrix} |$ $= 2 \hat{i} + 8 \hat{j} + \hat{k}$

and $\vec{c} = \hat{b} \times \hat{k} =$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 8 & 1 \\ 0 & 0 & 1 \end{matrix} |$ $= 8 \hat{i} - 2 \hat{j}$

$\Rightarrow \vec{c} - 2 \hat{j} = 8 \hat{i} - 4 \hat{j}$

Now, Projection of $\vec{c} - 2 \hat{j}$ on $\vec{a}$ is given by

$= (\vec{c} - 2 \hat{j}) \cdot \hat{a} = (8 i - 4 j) \cdot \frac{(3 i - j + 2 k)}{\sqrt{3^{2} + 1^{2} + 2^{2}}}$

$= \frac{24 + 4 + 0}{\sqrt{14}} = \frac{28}{\sqrt{14}} = 2 \sqrt{14}$ .

Solution

Q.
Let $\vec{a} = 3 \hat{i} - \hat{j} + 2 \hat{k}$ , $\vec{b} = \vec{a} \times (\hat{i} - 2 \hat{k})$ and $\vec{c} = \vec{b} \times \hat{k}$ . Then the projection of $\vec{c} - 2 \hat{j}$ on $\vec{a}$ is : [2025]

1 $2 \sqrt{7}$

2 $3 \sqrt{7}$

3 $2 \sqrt{14}$

4 $\sqrt{14}$

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Solution

Q. Let a→=3i^–j^+2k^, b→=a→×(i^–2k^) and c→=b→×k^. Then the projection of c→–2j^ on a→ is : [2025]

1 27

2 37

3 214

4 14