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← JEE Main 2022 · 27 Jun · Shift 2
MathematicsSingle correct+4 / -1Question 15 of 90

JEE Main 2022 · 27 Jun · Shift 2, Mathematics question 15

Let a⃗\vec a and b⃗\vec b be the vectors along the diagonals of a parallelogram having area 222\sqrt2. Let the angle between a⃗\vec a and b⃗\vec b be acute, ∣a⃗∣=1|\vec a|=1, and ∣a⃗⋅b⃗∣=∣a⃗×b⃗∣\left|\vec a\cdot\vec b\right|=\left|\vec a\times\vec b\right|. If c⃗=22(a⃗×b⃗)−2b⃗\vec c=2\sqrt2\left(\vec a\times\vec b\right)-2\vec b, then an angle between b⃗\vec b and c⃗\vec c is

Choose one option
Answer and worked solution

Official answer: D

Diagonals. Area =12∣a⃗×b⃗∣=22=\frac12|\vec a\times\vec b|=2\sqrt2, so ∣a⃗×b⃗∣=42|\vec a\times\vec b|=4\sqrt2; equal dot and cross magnitudes give a 45° angle, so ∣b⃗∣=8|\vec b|=8.

Vector c. b⃗⋅c⃗=−2∣b⃗∣2=−128\vec b\cdot\vec c=-2|\vec b|^2=-128 and ∣c⃗∣2=8⋅32+4⋅64=512|\vec c|^2=8\cdot32+4\cdot64=512.

Angle. cos⁡θ=−1288⋅162=−12\cos\theta=\dfrac{-128}{8\cdot16\sqrt2}=-\dfrac1{\sqrt2}, so θ=3π4\theta=\dfrac{3\pi}4.

(D)

Topics:Cross productDot product
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