First slide

Vectors

Question

Given that A + B + C = 0. Out of three vectors two are equal in magnitude and the magnitude of the third vector is $\sqrt{2}$ times that of either of the two having equal magnitude. Then the angles between vectors are given by

NA

A

30°,60°, 90°
B

45°,45°, 90°
C

45°,60°, 90°
D

90°, 135°135°

Solution

Let $| A | = | B | = a$
then $| C | = \sqrt{2} a$
Given that A+B+C = 0
or A+B = -C
Taking self Product, we have
$\begin{array}{l} (A + B) \cdot (A + B) = (- C) \cdot (- C) \\ A^{2} + B^{2} + 2 A \cdot B = C^{2} \\ a^{2} + a^{2} + 2 a^{2} \cos θ = 2 a^{2} \\ \cos θ = 0 or θ = 90^{\circ} \end{array}$
So, angle between A and B - 90"
Again B+C = -A
$\begin{array}{l} or (B + C) \cdot (B + C) = (- A) \cdot (- A) \\ or B^{2} + C^{2} + 2 B \cdot C = A^{2} \\ a^{2} + 2 a^{2} + 2 \sqrt{2} a^{2} \cos ϕ = a^{2} \\ \cos ϕ = - 1 / \sqrt{2} or ϕ = 135^{\circ} \end{array}$
So, angle between B and C is 135°'
similarly, angle between c and A is 135°.

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