Question

Find the sine of the angle between the vectors a=3i-4j+5k and b=i-j+k

Answer 1

The sine of angle between the vectors is √6/12.

• We have to find the sine of angle between the following vectors, a=3i-4j+5k and b=i-j+k

• The sine of the angle between vectors is given by

sinθ = $\frac{|\bar a \times \bar b|}{\bar a . \bar b}$

• We first find the value of,

$\bar a \times \bar b$ = $\left[\begin{array}{ccc}i&j&k\\3&-4&5\\1&-1&1\end{array}\right]$ = i(-4+5)-j(3-5)+k(-3+4)= i-2j+k

• Now, $|\bar a \times \bar b|$ = | i-2j+k| = $\sqrt{1^2+(-2)^2+1^2}$ = $\sqrt{1+4+1}$ = $\sqrt{6}$

• Now, $\bar a . \bar b[\tex] = (3i-4j+5k)(i-j+k) = 3+4+5 = 12</li></ul><ul><li>Now , sinθ = [tex]\frac{\sqrt{6}}{12}$