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← JEE Main 2014 · 19 Apr · Shift 1
MathematicsSingle correct+4 / -1Question 85 of 90

JEE Main 2014 · 19 Apr · Shift 1, Mathematics question 85

If x⃗=3i^−6j^−k^\vec x=3\hat i-6\hat j-\hat k, y⃗=i^+4j^−3k^\vec y=\hat i+4\hat j-3\hat k and z⃗=3i^−4j^−12k^\vec z=3\hat i-4\hat j-12\hat k, then the magnitude of the projection of x⃗×y⃗\vec x\times\vec y on z⃗\vec z is :

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Answer and worked solution

Official answer: C

x⃗×y⃗=∣i^j^k^3−6−114−3∣=22i^+8j^+18k^.\vec x\times\vec y=\begin{vmatrix}\hat i&\hat j&\hat k\\3&-6&-1\\1&4&-3\end{vmatrix}=22\hat i+8\hat j+18\hat k.
∣(22,8,18)⋅(3,−4,−12)∣13=∣66−32−216∣13=14.\frac{|(22,8,18)\cdot(3,-4,-12)|}{13}=\frac{|66-32-216|}{13}=14.