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CBSE Class 12 Math 2018 Solved Paper

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