Question

Find the possible unit vector parallel to the vector ‘$$x\hat{i}+y\hat{j}+z\hat{k}$$’, given that the vectors ‘$$a_1=x\hat{i}-\hat{j}+\hat{k}$$’ and ‘$$a_2=\hat{i}+y\hat{j}+z\hat{k}$$’ are collinear.
A. $$\left({\displaystyle \frac{1}{\sqrt{2}}}\right)\left(-\hat{j}+\hat{k}\right)$$
B. $$\left({\displaystyle \frac{1}{\sqrt{2}}}\right)\left(\hat{i}-\hat{j}\right)$$
C. $$\left({\displaystyle \frac{1}{\sqrt{3}}}\right)\left(\hat{i}-\hat{j}+\hat{k}\right)$$
D. $$\left({\displaystyle \frac{1}{\sqrt{3}}}\right)\left(\hat{i}+\hat{j}-\hat{k}\right)$$

Answer 1

Hint: Assume the proportionality of constants of unit vectors of the given vector to be a constant i.e. use the property of collinearity and then form a unit vector parallel to the given vector using different coefficient.

Formula used:
If two vectors $$x_1\hat{i}+y_1\hat{j}+z_1\hat{k}$$ and $$x_2\hat{i}+y_2\hat{j}+z_2\hat{k}$$ are parallel, then $${\displaystyle \frac{x_1}{x_2}}={\displaystyle \frac{y_1}{y_2}}={\displaystyle \frac{z_1}{z_2}}$$.
The unit vector of a vector $$x\hat{i}+y\hat{j}+z\hat{k}$$ is $${\displaystyle \frac{x\hat{i}+y\hat{j}+z\hat{k}}{\sqrt{x^2+y^2+z^2}}}$$

Complete step by step solution:
Given that, $$a_1=x\hat{i}-\hat{j}+\hat{k}$$ and $$a_2=\hat{i}+y\hat{j}+z\hat{k}$$.
Since $$a_1$$ and $$a_2$$ are colinear, thus the ratios of coefficient of $\hat{i},\hat{j},\hat{k}$ are equal.
$$\left({\displaystyle \frac{x}{1}}\right)={\displaystyle \frac{-1}{y}}={\displaystyle \frac{1}{z}}=\lambda$$
$$\Rightarrow x=\lambda$$, $$y=-{\displaystyle \frac{1}{\lambda }}$$ and $$z={\displaystyle \frac{1}{\lambda }}$$
Now putting the values of x,y, z in $$x\hat{i}+y\hat{j}+z\hat{k}$$
$$x\hat{i}+y\hat{j}+z\hat{k}=\left[\lambda \hat{i}-\left({\displaystyle \frac{1}{\lambda }}\right)\hat{j}+\left({\displaystyle \frac{1}{\lambda }}\right)\hat{k}\right]$$
Unit vector parallel to $$x\hat{i}+y\hat{j}+z\hat{k}$$
$$x\hat{i}+y\hat{j}+z\hat{k}=\pm {\displaystyle \frac{\left[\lambda \hat{i}-\left({\displaystyle \frac{1}{\lambda }}\right)\hat{j}+\left({\displaystyle \frac{1}{\lambda }}\right)\hat{k}\right]}{\sqrt{\left({\lambda }^2+\left({\displaystyle \frac{2}{{\lambda }^2}}\right)\right)}}}$$
Now, for $$\lambda =1$$
= $$\pm {\displaystyle \frac{\left[\hat{i}-\hat{j}+\hat{k}\right]}{\sqrt{3}}}$$

The correct answer is option C.

Note: Collinear vectors are those that are perpendicular to or located on a single line. Recalling that collinear vectors line up on the same or adjacent lines helps us begin the solution. We make use of the fact that one of the collinear vectors' components is equivalent to one vector's multiples. To demonstrate all of the collinear vector conditions, we use the fact that a collinear vector's cross product is zero.