The two vectors x2−1i^+(x+2)j^+x2k^ and 2i^−xj^+3k^ are orthogonal

For orthogonality, the scalar product= 0⇒2x2−1+(−x)(x+2)+3x2=0⇒2(2x+1)(x−1)=0⇒x=−12,1

The two vectors x2−1i^+(x+2)j^+x2k^ and 2i^−xj^+3k^ are orthogonal

For orthogonality, the scalar product= 0⇒2x2−1+(−x)(x+2)+3x2=0⇒2(2x+1)(x−1)=0⇒x=−12,1

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The two vectors $(x^{2} - 1) \hat{i} + (x + 2) \hat{j} + x^{2} \hat{k}$ and $2 \hat{i} - x \hat{j} + 3 \hat{k}$ are orthogonal

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for $x = \frac{1}{2}$

for $x = - \frac{1}{2}$ and $x = 1$

for no real value of x

for $x = - 1$

answer is D.

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Detailed Solution

For orthogonality, the scalar product= 0

$\begin{aligned} \Rightarrow 2 (x^{2} - 1) + (- x) (x + 2) + 3 x^{2} = 0 \\ \Rightarrow 2 (2 x + 1) (x - 1) = 0 \Rightarrow x = - \frac{1}{2}, 1 \end{aligned}$

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The two vectors x2−1i^+(x+2)j^+x2k^ and 2i^−xj^+3k^ are orthogonal

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The two vectors $(x^{2} - 1) \hat{i} + (x + 2) \hat{j} + x^{2} \hat{k}$ and $2 \hat{i} - x \hat{j} + 3 \hat{k}$ are orthogonal