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← JEE Main 2022 · 30 Jun · Shift 1
MathematicsSingle correct+4 / -1Question 17 of 90

JEE Main 2022 · 30 Jun · Shift 1, Mathematics question 17

Let a vector c⃗\vec c be coplanar with the vectors a⃗=−i^+j^+k^\vec a=-\hat i+\hat j+\hat k and b⃗=2i^+j^−k^\vec b=2\hat i+\hat j-\hat k. If the vector c⃗\vec c also satisfies the conditions c⃗⋅[(a⃗+b⃗)×(a⃗×b⃗)]=−42\vec c\cdot\left[\left(\vec a+\vec b\right)\times\left(\vec a\times\vec b\right)\right]=-42 and (c⃗×(a⃗−b⃗))⋅k^=3\left(\vec c\times\left(\vec a-\vec b\right)\right)\cdot\hat k=3, then the value of ∣c⃗∣2\left|\vec c\right|^2 is equal to :

Choose one option
Answer and worked solution

Official answer: C

Write c⃗=λa⃗+μb⃗\vec c=\lambda\vec a+\mu\vec b. (a⃗+b⃗)×(a⃗×b⃗)=4a⃗−b⃗(\vec a+\vec b)\times(\vec a\times\vec b)=4\vec a-\vec b, and c⃗×(a⃗−b⃗)=−(λ+μ)a⃗×b⃗\vec c\times(\vec a-\vec b)=-(\lambda+\mu)\vec a\times\vec b with a⃗×b⃗=(−2,1,−3)\vec a\times\vec b=(-2,1,-3).

Conditions. λ+μ=1\lambda+\mu=1 and 14λ−14μ=−4214\lambda-14\mu=-42: λ=−1\lambda=-1, μ=2\mu=2, c⃗=(5,1,−3)\vec c=(5,1,-3).

∣c⃗∣2=35|\vec c|^2=35.

(C)

Topics:Scalar and vector triple productsCross product
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