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
B
23
C
2
D
43

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 $$x = 2, \displaystyle - \dfrac{2}{3}$$Maths

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