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Vector Algebra Part 1

Section: Mathematics

Question : 14 of 100

Marks: +1, -0

If vectors

a_{1}=x\hat{i}-\hat{j}+\hat{k}

a_{2}=\hat{i}+\hat{y}j+z\hat{k}

x\hat{i}+\hat{j}+z\hat{k}

Solution:

Given,

a_{1}=x\hat{i}-\hat{j}+\hat{k}

and

a_{2}=\hat{i}+y\hat{j}+z\hat{k}

are collinear, then

\;\frac{x}{1}=\;\frac{-1}{y}=\;\frac{1}{z}=\lambda

(Say)

This gives

x=\lambda, y=-\;\frac{1}{\lambda}, z=\;\frac{1}{\lambda}

Then, unit vector parallel to vector

x\hat{i}+y\hat{j}+

zk will be

\;\; =\;\frac{\pm\left((\lambda)\hat{i}-\left(\frac{1}{\lambda}\right)\hat{j}+\left(\frac{1}{\lambda}\right)\hat{k}\right)}{\sqrt{(\lambda)^{2}+\left(\; \frac{-1}{\lambda}\right)^{2}+\left(\; \frac{1}{\lambda}\right)^{2}}}

\;\; =\pm\;\frac{(\lambda^{2}\hat{i}-\hat{j}+\hat{k})\lambda}{\lambda\sqrt{\lambda^{4}+2}}=\pm\;\frac{\lambda^{2}\hat{i}-\hat{j}+\hat{k}}{\sqrt{\lambda^{4}+2}}

Take,

\lambda=1=\frac{\pm(\hat{i}-\hat{j}+\hat{k})}{\sqrt{3}}

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