Find |x|, if for a unit vector a, (x - a). (x + a) = 12.

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Solution

Given, |a| = 1

( $∵$ It is a unit vector, magnitude of a unit vector is 1)

and $(x - a) . (x + a) = 12$

$\Rightarrow x . x + x . a - a . x - a . a = 12$ $(∵ a . a = | a |^{2} a n d a . b = b . a)$

$\Rightarrow | x |^{2} - | a |^{2} = 12 \Rightarrow | x |^{2} - 1^{2} = 12$ [ $∵ | a | = 1$ as a is a unit vector]

$\Rightarrow | x |^{2} = 13 \Rightarrow | x | = \sqrt{13}$

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Similar questions

If $\to a$ is unit vector and $(\to x - \to a) . (\to x + \to a) = 8, t h e n | \to x | =$

| \to x |

\to a, (\to x - \to a) . (\to x + \to a) = 15.

| \to x |

\to a, (\to x - \to a) \cdot (\to x + \to a) = 12

\to x, \to y, \to z

\to x + \to y + \to z = \to a

(\to x \times \to y) \times \to z = \to b

\to a \cdot \to x = \frac{3}{2}, \to a \cdot \to y = \frac{7}{4}

| \to a | = 2,

\vec{a}

|\vec{x}|

(\vec{x} - \vec{a}) \cdot (\vec{x} + \vec{a}) = 8

(\vec{x} - \vec{a}) \cdot (\vec{x} + \vec{a}) = 12

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