The vectors a→=xi^+2j^+5k^ and b→=i^+yj^−zk^ are collinear, if

Since, a→ and b→ are collinear.∴a→=λb→⇒(xi^−2j^+5k^)=λ(i^+yj^−zk^)On comparingx=λ,−2=λy and 5=−λzFor λ=1x=1,y=−2 and z=−5 For λ=1/2x=1/2,y=−4 and z=−10 For λ=−1/2x=−1/2,y=4 and z=−10all options are correct.

The vectors a→=xi^+2j^+5k^ and b→=i^+yj^−zk^ are collinear, if

Since, a→ and b→ are collinear.∴a→=λb→⇒(xi^−2j^+5k^)=λ(i^+yj^−zk^)On comparingx=λ,−2=λy and 5=−λzFor λ=1x=1,y=−2 and z=−5 For λ=1/2x=1/2,y=−4 and z=−10 For λ=−1/2x=−1/2,y=4 and z=−10all options are correct.

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The vectors $\vec{a} = x \hat{i} + 2 \hat{j} + 5 \hat{k} and \vec{b} = \hat{i} + y \hat{j} - z \hat{k}$ are collinear, if

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$x = 1 / 2, y = - 4, z = - 10$

All of these

$x = - 1 / 2, y = 4, z = 10$

$x = 1, y = - 2, z = - 5$

answer is D.

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Detailed Solution

Since, $\vec{a} and \vec{b}$ are collinear.

$\begin{array}{l} ∴ \vec{a} = λ \vec{b} \\ \Rightarrow (x \hat{i} - 2 \hat{j} + 5 \hat{k}) = λ (\hat{i} + y \hat{j} - z \hat{k}) \end{array}$

On comparing

$x = λ, - 2 = λ y and 5 = - λ z$

For $λ = 1$

$\begin{aligned} x = 1, y = - 2 and z = - 5 \\ For λ = 1 / 2 \\ x = 1 / 2, y = - 4 and z = - 10 \\ For λ = - 1 / 2 \\ x = - 1 / 2, y = 4 and z = - 10 \end{aligned}$