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Question

The vector $$\widehat{i} + x \widehat{j} + 3 \widehat{k} $$ is rotated through an angle $$\theta$$ and is doubled in magnitude. It now becomes $$4\widehat{i}+(4x - 2)\widehat{j} + 2\widehat{k}$$. The value(s) of $$x$$ are


A
1
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B
23
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C
2
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D
43
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Solution

The correct options are
B $$\displaystyle \frac{-2}{3}$$
C $$2$$
Let $$\vec{\alpha} = \widehat{i} + x \widehat{j} + 3\widehat{k}, \vec{\beta} = 4 \widehat{i} + (4x - 2)\widehat{j} + 2 \widehat{k}$$
Given, $$2 |\vec{\alpha}| = |\vec{\beta}|$$
or $$2 \sqrt{10 + x^2} = \sqrt{20 + 4 (2x - 1)^2}$$
or $$10 + x^2 = 5 + (4x^2 - 4x + 1)$$
or $$3x^2 - 4x - 4 = 0$$
or $$x = 2, \displaystyle - \dfrac{2}{3}$$

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