Question

Show that vectors a vector = 2i +3j + 6k b vector = 3i - 6j + 2k and c vector = 6i + 2j - 3k are mutually perpendicular.​

Answer 1

Answer:

In order to prove that the two vectors are perpendicular, their dot product should be zero which has been shown.

Explanation:

Given:

$\vec a = 2\hat i + 3\hat j+ 6\hat k \\\vec b = 3\hat i -6\hat j + 2\hat k\\\vec c = 6\hat i + 2\hat j -3\hat k$

The dot product of two vectors $\vec A$ and $\vec B$ is given by

$\vec A \cdot \vec B = AB \cos \theta$

These two vectors are perpendicular when $\theta = 90 ^\circ$ i.e.,  their dot product is 0.

we know

$\hat i\cdot \hat i=\hat j\cdot \hat j=\hat k\cdot \hat k = 1\\\hat i\cdot \hat j=\hat j\cdot \hat k=\hat k\cdot \hat i=0.$

Therefore,

$\vec a \cdot \vec b = (2\hat i + 3\hat j + 6\hat k)\cdot (3\hat i - 6\hat j+ 2\hat k)\\ = (2\times 3)+ (3\times -6)+(6\times 2) \\= 6-18+12\\ = 0.$

$\vec b \cdot \vec c = (3\hat i - 6\hat j+2\hat k)\cdot (6\hat i+ 2\hat j-3\hat k)\\ = (3\times 6)+(-6\times 2)+(2\times -3)\\ = 18-2-6\\ = 0.$

$\vec a \cdot \vec c = (2\hat i+3\hat j + 6\hat k)\cdot (6\hat i+2\hat j-3\hat k)\\=(2\times 6)+(3\times 2)+(6\times -3)\\=12+6-18\\=0.$

Since, dot products of all three vector with each other are zero therefore all these vectors are mutually perpendicular.