For two vectors {P} and {Q}, {P} · {Q}=|{P} × {Q}| The magnitude of {R}={P}+{Q} is ( 45^{}={1}{{2}} ) ?

Correct Answer is : P2+Q2+2PQ{P²+Q²+{2} P Q}P2+Q2+2PQ

Correct Answer is : P2+Q2+2PQ{P²+Q²+{2} P Q}P2+Q2+2PQ

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MHTCET 19 Apr 2026 Shift 1 Paper

Question : 1 of 150

Marks: +1, -0

\vec{P}

\vec{Q}, \vec{P} \cdot \vec{Q}=|\vec{P} \times \vec{Q}|

\vec{R}=\vec{P}+\vec{Q}

\cos 45^{\circ}=\frac{1}{\sqrt{2}}

) ?

Solution:

Concept:

The dot product equals the magnitude of the cross product, so the angle between the vectors is

45^\circ

Explanation:

Let

\theta

be the angle between

\vec{P}

and

\vec{Q}

\vec{P} \cdot \vec{Q} = PQ\cos\theta

|\vec{P} \times \vec{Q}| = PQ\sin\theta

Given:

PQ\cos\theta = PQ\sin\theta

So,

\cos\theta = \sin\theta

, which gives

\theta = 45^\circ

For

\vec{R} = \vec{P} + \vec{Q}

R = \sqrt{P^2 + Q^2 + 2PQ\cos\theta}

Substitute

\theta = 45^\circ

and

\cos 45^\circ = \frac{1}{\sqrt{2}}

R = \sqrt{P^2 + Q^2 + 2PQ \cdot \frac{1}{\sqrt{2}}}

R = \sqrt{P^2 + Q^2 + \sqrt{2}PQ}

Answer:

\sqrt{P^{2}+Q^{2}+\sqrt{2}PQ}

Option D.

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