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CBSE Class 12 Math 2018 Solved Paper

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Question : 20 of 29
Marks: +1, -0
Let a⃗\vec{a} = 4i^+5j^−k^4\hat{i}+5\hat{j}-\hat{k} , b⃗\vec{b} = i^−4j^+5k^\hat{i}-4\hat{j}+5\hat{k} and c⃗\vec{c} = 3i^+j^−k^3\hat{i}+\hat{j}-\hat{k}. Find a vector d⃗\vec{d} which is perpendicular to both c⃗\vec{c} and b⃗\vec{b} and d⃗⋅a⃗\vec{d}\cdot\vec{a} = 21.
Solution:  
Let d⃗\vec{d} = xi^+yj^+zk^x\hat{i}+y\hat{j}+z\hat{k}
since d⃗\vec{d} is perpendicular to c⃗\vec{c} and b⃗\vec{b} , so their dot product is zero
d⃗⋅c⃗\vec{d}\cdot\vec{c} = 0 and d⃗⋅b⃗\vec{d}\cdot\vec{b} = 0
3x + y - z = 0 ... (1)
x - 4y + 5z = 0 ... (2)
Also given that d⃗⋅a⃗\vec{d}\cdot\vec{a} = 21.
4x + 5y - z = 21 ... (3)
Solving all the equations simultaneously
x = −13-\frac{1}{3} , y = 163\frac{16}{3} , z = 133\frac{13}{3}
d⃗\vec{d} = 13\frac{1}{3} (−i^+16j^+13k^)(-\hat{i}+16\hat{j}+13\hat{k})
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