Question

* Find the angle between
a= 21+3j +k. And. b= -21-2j-2k.
OR
Prove that a= 2i+3j+k. And b= 2i+4j+8k are mutually perpendicular.​

Answer 1

Explanation:

The vectors,

a = 2i + 3j + k

b = -2i - 2j - 2k

Let the angle between a vector and b vector be Alpha

Angle between a and b:

$\cos( \alpha ) = \frac{a.b}{ |a| |b| }$

a.b (dot product) = - 4 - 6 - 2 = -12

|a| and |b| are the magnitude of a vector and b vector

$|a| = \sqrt{ {2}^{2} + {3}^{2} + {1}^{2} } = \sqrt{14}$

$|b| = \sqrt{ { - 2}^{2} + { - 2}^{2} - { - 2}^{2} } = \sqrt{12}$

$|a| \times |b| = \sqrt{14} \sqrt{12} = 168$

Substitute the values in the formula,

$\cos( \alpha ) = \frac{ - 12}{168} = - \frac{1}{14}$

(OR)

The given vectors are,

a = 2i + 3j + k

b = 2i + 4j + 8k

To prove: The vectors are perpendicular

As per the rule, if the dot product of the vector is zero, then the vectors are perpendicular

a.b = 4 + 12 + 8 = 24 ≠ 0

So, the vectors are not perpendicular