Question

If vectors \(\vec{a_1} = x\hat{i} - \hat{j} + \hat{k}\) and \(\vec{a_2} = \hat{i} + y\hat{j} + z\hat{k}\) are collinear, then a possible unit vector parallel to the vector \(x\hat{i} + y\hat{j} + z\hat{k}\) is :

• A. \(\frac{1}{\sqrt3}\left(\hat{i} - \hat{j} + \hat{k}\right)\)
• B. \(\frac{1}{\sqrt2}\left(-\hat{j} + \hat{k}\right)\)
• C. \(\frac{1}{\sqrt2}\left(\hat{i} - \hat{j}\right)\)
• D. \(\frac{1}{\sqrt3}\left(\hat{i} + \hat{j} - \hat{k}\right)\)

Hint:

Collinear means proportional components: \(\frac{x}{1} = \frac{-1}{y} = \frac{1}{z}\); so \(xz = 1\) and \(y = -z\).

Answer 1

The Correct Option is A

Collinear vectors have proportional components: \(\frac{x}{1} = \frac{-1}{y} = \frac{1}{z} = k\).
So \(x = k\), \(y = -\frac1k\), \(z = \frac1k\) (with \(k \ne 0\)).
The vector \(x\hat{i} + y\hat{j} + z\hat{k} = k\hat{i} - \frac1k\hat{j} + \frac1k\hat{k}\). For \(k = 1\) it is \(\hat{i} - \hat{j} + \hat{k}\), of magnitude \(\sqrt3\); the unit vector is \(\frac{1}{\sqrt3}(\hat{i} - \hat{j} + \hat{k})\).
The other options cannot occur: (2) has \(x = 0\), impossible since \(x = k \ne 0\); (3) has \(z = 0\), impossible since \(z = \frac1k \ne 0\); (4) has components in ratio \(1 : 1 : -1\), but \(y = -z\) and \(xz = 1\) require \(x\) and \(z\) of the same sign with \(y\) opposite to \(z\), so \(y = +x\) is impossible.
Answer: (1) \(\frac{1}{\sqrt3}(\hat{i} - \hat{j} + \hat{k})\)