Let a = 6 i ^ + 9 j ^ + 12 k ^ , b = i ^ + 11 j ^ - 2 k ^ and c be vectors such that a × c = a × b . If a · c = - 12 , c· ( i ^ - 2 j ^ + k ^ ) = 5 , then c· ( i ^ + j ^ + k ^ ) is equal to _______. [2023]

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Solution

Q.
Let $\vec{a} = 6 \hat{i} + 9 \hat{j} + 12 \hat{k},$ $\vec{b} = α \hat{i} + 11 \hat{j} - 2 \hat{k}$ and $\vec{c}$ be vectors such that $\vec{a} \times \vec{c} = \vec{a} \times \vec{b} .$ If $\vec{a} \cdot \vec{c} = - 12,$ $\vec{c} \cdot (\hat{i} - 2 \hat{j} + \hat{k}) = 5,$ then $\vec{c} \cdot (\hat{i} + \hat{j} + \hat{k})$ is equal to _______. [2023]

Ans.
(11)

$We have, \vec{a} = 6 \hat{i} + 9 \hat{j} + 12 \hat{k}$

$\vec{b} = α \hat{i} + 11 \hat{j} - 2 \hat{k}$

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

$∵$ $\vec{a} \cdot \vec{c} = - 12$

$∴$ $6 x + 9 y + 12 z = - 12$ ...(i)

Also, $\vec{c} \cdot (\hat{i} - 2 \hat{j} + \hat{k}) = 5$

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

Now, $\vec{a} \times \vec{c} = \vec{a} \times \vec{b}$

$\Rightarrow \hat{i} (9 z - 12 y) - \hat{j} (6 z - 12 x) + \hat{k} (6 y - 9 x)$ $= - 150 \hat{i} - \hat{j} (- 12 - 12 α) + \hat{k} (66 - 9 α)$

$Comparing both sides, we get$

$9 z - 12 y = - 150$ ...(iii)

$Solving (i) and (ii), we get$

$21 y + 6 z = - 42$ ...(iv)

$Now, solving (iii) and (iv), we get y = 2 and z = - 14$

$Put these values in (ii), we get x = 23$

$\Rightarrow \vec{c} = 23 \hat{i} + 2 \hat{j} - 14 \hat{k}$

$∴$ $\vec{c} \cdot (\hat{i} + \hat{j} + \hat{k}) = 23 + 2 - 14 = 11$

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