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If vectors ${\vec{a}}_{1} = x \hat{i} - \hat{j} + \hat{k} a n d {\vec{a}}_{2} = \hat{i} + y \hat{j} + z \hat{k}$ are collinear, then a possible unit vector parallel to the vector $x \hat{i} + y \hat{j} + z \hat{k}$ is:

Option 1 -

$\frac{1}{\sqrt[]{3}} (\hat{i} + \hat{j} - \hat{k})$

Option 2 -

$\frac{1}{\sqrt[]{2}} (- \hat{j} + \hat{k})$

Option 3 -

$\frac{1}{\sqrt[]{3}} (\hat{i} - \hat{j} + \hat{k})$

Option 4 -

$\frac{1}{\sqrt[]{2}} (\hat{i} - \hat{j})$

3 Views | Posted 2 weeks ago

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1 Answer

Answered by

alok kumar singh | Contributor-Level 10

2 weeks ago

Correct Option - 3

Detailed Solution:

${\vec{a}}_{1} = x \hat{i} - \hat{j} + \hat{k} & {\vec{a}}_{2} = \hat{i} + y \hat{j} + z \hat{k}$

given ${\vec{a}}_{1} & {\vec{a}}_{2}$ are collinear then ${\vec{a}}_{1} = λ {\vec{a}}_{2}$

$\Rightarrow (x \hat{i} - \hat{j} + \hat{k}) = λ (\hat{i} + y \hat{j} + z \hat{k})$

Since $\hat{i}, \hat{j} & \hat{k}$ are not collinear so

$S o x \hat{i} + y \hat{j} + z \hat{k} = λ \hat{i} - \frac{1}{λ} \hat{j} + \frac{1}{λ} \hat{k}$

Hence possible unit vector parallel to it be $\frac{1}{\sqrt[]{3}} (\hat{i} - \hat{j} + \hat{k})$ for $λ$ =

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If vectors ${\vec{a}}_{1} = x \hat{i} - \hat{j} + \hat{k} a n d {\vec{a}}_{2} = \hat{i} + y \hat{j} + z \hat{k}$ are collinear, then a possible unit vector parallel to the vector $x \hat{i} + y \hat{j} + z \hat{k}$ is:

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If vectors a → 1 = x i ^ − j ^ + k ^ a n d a → 2 = i ^ + y j ^ + z k ^ are collinear, then a possible unit vector parallel to the vector x i ^ + y j ^ + z k ^ is:

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If vectors ${\vec{a}}_{1} = x \hat{i} - \hat{j} + \hat{k} a n d {\vec{a}}_{2} = \hat{i} + y \hat{j} + z \hat{k}$ are collinear, then a possible unit vector parallel to the vector $x \hat{i} + y \hat{j} + z \hat{k}$ is: