First slide

Vectors

Question

Given that A + B = C and that C is perpendicular to A. Further $| A | = | C |$ then what is the angle between A and B?

Moderate

A

$\frac{π}{4} radian$
B

$\frac{π}{2} radian .$
C

$\frac{3 π}{4} radian$
D

$π radian$

Solution

Since C is perpendicular to A, therefore,
A² +C² = B²
But A=C. Hence $C = B / \sqrt{2} = A$
so,
$\begin{array}{l} \cos θ = \frac{C^{2} - A^{2} - B^{2}}{2 AB} = \frac{(B^{2} / 2) - (B^{2} / 2) - B^{2}}{2 (B / \sqrt{2}) B} \\ = \frac{- B^{2}}{\sqrt{2} B^{2}} = - \frac{1}{\sqrt{2}} \\ θ = 135^{\circ} or (3 π / 4) rad \end{array}$
See figure

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