Let a = i ^ + 4 j ^ + 2 k ^ , b = 3 i ^ - 2 j ^ + 7 k ^ and c = 2 i ^ - j ^ + 4 k ^ . If a vector d satisfies d × b = c × b and d · a = 24 , then | d | 2 is equal to [2023]

Question

Let $\vec{a} = \hat{i} + 4 \hat{j} + 2 \hat{k}$ , $\vec{b} = 3 \hat{i} - 2 \hat{j} + 7 \hat{k}$ and $\vec{c} = 2 \hat{i} - \hat{j} + 4 \hat{k}$ . If a vector $\vec{d}$ satisfies $\vec{d} \times \vec{b} = \vec{c} \times \vec{b}$ and $\vec{d} \cdot \vec{a} = 24$ , then $| \vec{d} |^{2}$ is equal to [2023]

Smart Achievers · Accepted Answer

(3)

We have, $\vec{a} = \hat{i} + 4 \hat{j} + 2 \hat{k}, \vec{b} = 3 \hat{i} - 2 \hat{j} + 7 \hat{k}$ and $\vec{c} = 2 \hat{i} - \hat{j} + 4 \hat{k}$

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

Now, $\vec{c} \times \vec{b} =$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & - 1 & 4 \\ 3 & - 2 & 7 \end{matrix} |$

$= \hat{i} (- 7 + 8) - \hat{j} (14 - 12) + \hat{k} (- 4 + 3) = \hat{i} - 2 \hat{j} - \hat{k}$

Now, $\vec{d} \times \vec{b} =$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ x & y & z \\ 3 & - 2 & 7 \end{matrix} |$

$= \hat{i} (7 y + 2 z) - \hat{j} (7 x - 3 z) + \hat{k} (- 2 x - 3 y)$

We have, $\vec{d} \times \vec{b} = \vec{c} \times \vec{b}$

$∴$ $\hat{i} (7 y + 2 z) - \hat{j} (7 x - 3 z) + \hat{k} (- 2 x - 3 y) = \hat{i} - 2 \hat{j} - \hat{k}$

$7 y + 2 z = 1 \dots (i)$

$7 x - 3 z = 2 \dots (i i)$

$2 x + 3 y = 1 \dots (i i i)$

$\vec{d} \cdot \vec{a} = (x \hat{i} + y \hat{j} + z \hat{k}) \cdot (\hat{i} + 4 \hat{j} + 2 \hat{k}) = 24$

$\Rightarrow x + 4 y + 2 z = 24 \dots (i v)$

By solving (i) and (iv), we get

$x - 3 y = 23 \dots (v)$

By solving (iii) and (v), we get $x = 8$ and $y = - 5$

Substitute the value of $x$ in (ii), we get $7 (8) - 3 z = 2 \Rightarrow z = 18$

$∴$ $\vec{d} = 8 \hat{i} - 5 \hat{j} + 18 \hat{k}$ $\Rightarrow | \vec{d} |^{2} = 8^{2} + {(- 5)}^{2} + 18^{2} = 413$

Solution

1 423

2 313

3 413

4 323

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