Question

Two vectors p and q such that |p+q|=|p -q|.then the angle between the vectors will be

Answer 1

Answer:

Angle between the vectors is 90°.

Explanation:

Let the angle between them be A.

Using the properties of vectors :

• Resultant of two vectors( A and B, and C as angle b/w them ) = √( |A|² + |B|² + 2|A||B|cosC )

Here,

= > | p + q | = | p - q |

= > √( p² + q² + 2pq.cosA ) = √( p² + q² - 2pq.cosA ) { p & q are now representing the magnitudes }

= > p² + q² + 2pq.cosA = p² + q² - 2pq.cosA

= > 2pq.cosA = - 2pq.cosA

= > 2pq.cosA + 2pq.cosA = 0

= > 4pq.cosA = 0

= > cosA = 0 { 4, p, q ≠ 0 }

= > cosA = cos90°

Therefore angle between these vectors is 90°

Answer 2

$\huge{\underline{\underline{\red{Answer}}}}$

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$\large{\mathbb{\blue{GIVEN}}}$

|$\overrightarrow{P}$+$\overrightarrow{Q}$|=|$\overrightarrow{P}-$$\overrightarrow{Q}$|

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$\huge{\mathfrak{\red{Solution}}}$

|$\overrightarrow{P}$+$\overrightarrow{Q}$|=|$\overrightarrow{P}$-$\overrightarrow{Q}$|

Squaring on both side,

${P}^{2} + 2P.Q + {Q}^{2} = {P}^{2} - 2P.Q+ {Q}^{2} \\ \implies {P}^{2} -{P}^{2} +{Q}^{2} -{Q}^{2} = 2P.Q+ 2P.Q \\\implies 4P.Q = 0 \\ \implies P.Q = 0 \\ \implies PQ\cos \theta = 0 \\ \implies \cos \theta = 0 \\ \implies \cos \theta = \cos90°\\ \implies \theta = 90$

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${\mathfrak{\bold{\red{The\:Angel\:is\:90}}}}$