Mathematics in Physics

JEE Main level, numerical answer, medium

45 of 121

Question 45 of 121

Vectors ai^+bj^+k^a\hat{i}+b\hat{j}+\hat{k} and 2i^−3j^+4k^2\hat{i}-3\hat{j}+4\hat{k} are perpendicular to each other when 3a+2b=73a+2b=7, the ratio of aa to bb is x2\dfrac{x}{2}. The value of xx is _____.

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Answer: 1

Dot product of two perpendicular vectors will be zero. Therefore,
(ai^+bj^+k^)⋅(2i^−3j^+4k^)=0\left(a\hat{i}+b\hat{j}+\hat{k}\right)\cdot \left(2\hat{i}-3\hat{j}+4\hat{k}\right)=0
⇒2a−3b+4=0...Eq(1)\Rightarrow 2a-3b+4=0...Eq(1)
As given in the question,
3a+2b=7...Eq(2)3a+2b=7...Eq(2)
From both equations, we get a=1&b=2
Then,
ab=x2⇒x=2ab=2×12\dfrac{a}{b}=\dfrac{x}{2} \Rightarrow x=\dfrac{2a}{b}=\dfrac{2\times 1}{2}
⇒x=1\Rightarrow x=1

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