First slide

Vectors

Question

Given that P + Q = R and that R is perpendicular to P. If IPI = IRl, then what is the angle between P and a ?

Moderate

A

$π / 4 rad$
B

$3 π / 4 rad$
C

$π / 2 rad$
D

$πrad$

Solution

Given P+Q=R or R-P=Q
Since R is perpendicular to P, therefore,
R² + P² = Q²
But P=R. Hence $R = Q / \sqrt{2} = P$
So, $\cos θ = \frac{R^{2} - P^{2} - Q^{2}}{2 PQ} = \frac{\frac{Q^{2}}{2} - \frac{Q^{2}}{2} - Q^{2}}{2 \frac{Q}{\sqrt{2}} (Q)}$
$\begin{array}{l} = \frac{- Q^{2}}{\sqrt{2} Q^{2}} = - \frac{1}{\sqrt{2}} \\ θ = 135^{\circ} or 3 π / 4 rad \end{array}$

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