Question

35. Show that vectors A = 2i - 3j- k and B = 6i+ 9j + 3k are parallel.​

Answer 1

Solution :

$\sf{a = 2 \hat{i} - 3 \hat{j} - \hat{k}} \\ \\ \longrightarrow \: \sf{ |a| = \sqrt{(2) {}^{2} + ( - 3) {}^{2} + ( - 1) {}^{2} } } \\ \\ \longrightarrow \: \sf{ |a| = \sqrt{14} \:}$

Also,

$\sf{b = 6 \hat{i} + 9 \hat{j} + 3 \hat{k}} \\ \\ \longrightarrow \: \sf{ |b| = \sqrt{ {6}^{2} + {9}^{2} + {3}^{2} } } \\ \\ \longrightarrow \: \sf{ |b| = 3 \sqrt{14} }$

DoT ProducT

$\sf{ \hat{ {i}}^{2} = \hat{{j}^{2}} = \hat{k}^{2} = 0} \\ \sf{\hat{i} \hat{j} = \hat{j} \hat{k} = \hat{i} \hat{k} = 1}$

$\sf{a \times b = |a| |b| \cos( \alpha ) } \\ \\ \implies \: \sf{12 - 27 - 3 = \sqrt{14} \times 3 \sqrt{14} \times \cos( \alpha ) } \\ \\ \implies \: \sf{ 0 = 3( \sqrt{14}) {}^{2} \times \cos( \alpha ) } \\ \\ \implies \: \sf{ \cos( \alpha ) = 0 }$

• For two vectors to be parallel,the angle in between them should be 0°

Thus,the given vectors are parallel to eachother

Answer 2

Question:- Show that vectors A = 2i -3j -k and B = -6i + 9j + 3k are parallel

Explanation:

Given,

A = 2i - 3j - k

B = -6j + 9j + 3k

Two vectors are said to be parallel if there cross product is null vector.

Thus we have to show that,

$\vec{A}\times\vec{B}=\vec{0}$

Now consider,

$\vec{A}\times\vec{B}=(2\hat{i}-3\hat{j}-\hat{k})\times(-6\hat{i}+9\hat{j}+3\hat{k})$

$\vec{A}\times\vec{B}={\left|\begin{array}{ccc}{\hat{i}}&{\hat{j}}&{\hat{k}}\\2&{-3}&{-1}\\{-6}&9&3\end{array}\right|}$

$\vec{A}\times\vec{B}=(-9+9)\hat{i}-(6-6)\hat{j}+(18-18)\hat{k}$

$\vec{A}\times\vec{B}=0\hat{i}-0\hat{j}+0\hat{k}$

$\vec{A}\times\vec{B}=\vec{0}$

Thus vector A and vector B are parallel.