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← JEE Main 2022 · 26 Jul · Shift 2
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JEE Main 2022 · 26 Jul · Shift 2, Mathematics question 15

A vector a⃗\vec a is parallel to the line of intersection of the plane determined by the vectors i^\hat i, i^+j^\hat i+\hat j and the plane determined by the vectors i^−j^\hat i-\hat j, i^+k^\hat i+\hat k. The obtuse angle between a⃗\vec a and the vector b⃗=i^−2j^+2k^\vec b=\hat i-2\hat j+2\hat k is

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Answer and worked solution

Official answer: A

Normals. i^×(i^+j^)=k^\hat i\times(\hat i+\hat j)=\hat k and (i^−j^)×(i^+k^)=−i^−j^+k^(\hat i-\hat j)\times(\hat i+\hat k)=-\hat i-\hat j+\hat k.

Line of intersection ∥k^×(−i^−j^+k^)=i^−j^\parallel\hat k\times(-\hat i-\hat j+\hat k)=\hat i-\hat j.

Angle with b⃗\vec b. cos⁡θ=±332\cos\theta=\pm\frac{3}{3\sqrt2}; the obtuse one is 3π4\frac{3\pi}4.

(A)

Topics:Cross productDot productEquation of a plane
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