Vector in three dimensions | Part -3

Vector Operations: Questions and Solutions

i) Given vector \( \vec{a} = 2 \hat{i} - 3 \hat{j} + 6 \hat{k} \)

Solution:

Direction ratios are: 2, -3, 6

Magnitude = \( \sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 \)

Direction cosines are: \( \frac{2}{7}, \frac{-3}{7}, \frac{6}{7} \)

ii) Let vectors \( \vec{a} = 2 \hat{i} - 3 \hat{j} + 4 \hat{k} \) and \( \vec{b} = m \hat{i} + 6 \hat{j} - 8 \hat{k} \)

Solution:

For collinearity, \( \frac{2}{m} = \frac{-3}{6} = \frac{4}{-8} \Rightarrow \frac{2}{m} = -\frac{1}{2} \Rightarrow m = -4 \)

iii) \( \vec{a} = 2 \hat{i} + 3 \hat{j} - \hat{k} \), \( \vec{b} = \hat{i} - 2 \hat{j} + \hat{k} \)

Solution:

Resultant \( \vec{R} = (2+1)\hat{i} + (3-2)\hat{j} + (-1+1)\hat{k} = 3\hat{i} + 1\hat{j} \)

Magnitude of \( \vec{R} = \sqrt{3^2 + 1^2} = \sqrt{9 + 1} = \sqrt{10} \)

Unit vector in direction of \( \vec{R} = \frac{1}{\sqrt{10}}(3\hat{i} + 1\hat{j}) \)

Vector of magnitude 5 in same direction = \( 5 \times \frac{1}{\sqrt{10}}(3\hat{i} + 1\hat{j}) = \frac{15}{\sqrt{10}}\hat{i} + \frac{5}{\sqrt{10}}\hat{j} \)