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Match the LIST-I with LIST-II

LIST-ILIST-II
A. Area of triangle determined by vectors (i^+j^)(\hat{i}+\hat{j}) and (i^−2j^+k^)(\hat{i}-2\hat{j}+\hat{k}) isI. 0
B. The value of (i^×j^).j^+(j^×i^).k^+(k^×i^).j^(\hat{i}\times\hat{j}).\hat{j} + (\hat{j}\times\hat{i}).\hat{k} + (\hat{k}\times\hat{i}).\hat{j} isII. 112\dfrac{\sqrt{11}}{2}
C. If (2λ−1)i^+3j^−k^(2\lambda-1)\hat{i}+3\hat{j}-\hat{k} and i^+λj^+2k^\hat{i}+\lambda\hat{j}+2\hat{k} are perpendicular to each other then λ\lambda isIII. 4
D. Projection of 5i^+j^+4k^5\hat{i}+\hat{j}+4\hat{k} on 2i^+6j^+3k^2\hat{i}+6\hat{j}+3\hat{k} isIV. 3/5

Choose the correct answer from the options given below:

Solution

✅ Correct Option: 1

A: cross product =(1,−1,−3)=(1,-1,-3), magnitude 11\sqrt{11}, so area =112=\dfrac{\sqrt{11}}{2} (II).

B: 0+(−1)+1=00+(-1)+1=0 (I).

C: (2λ−1)+3λ−2=0(2\lambda-1)+3\lambda-2=0 gives λ=35\lambda=\dfrac35 (IV).

D: 10+6+127=4\dfrac{10+6+12}{7}=4 (III).

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