6) Angle between vector \( \vec{A}=\vec{\imath}+2 \vec{\jmath}-\vec{k} \) and \( \vec{B}=-\vec{\imath}-2 \vec{\jmath}+\vec{k} \) will be.. a)zero \( \begin{array}{ll}\text { b) } 45^{\circ} & \text { c) } 90^{\circ}\end{array} \) d) \( 180^{\circ} \)
Call
Given \(\vec A\) = \(\hat i+2\hat j-\hat k\)
\(\vec B\) = \(-\hat i-2\hat j+\hat k\)
Angle b/w \(\vec A\) and \(\vec B\)
cos θ = \(\frac{\vec A.\vec B}{|\vec A|.|\vec B|}\)
\(=\frac{(\hat i+2\hat j-\hat k).(-\hat i-2\hat j+\hat k)}{\sqrt{1^2+(2)^2+(1)^2}\sqrt{(-1)^2+(-2)^2+(1)^2}}\)
cos θ = \(\frac{-1-4-1}{\sqrt6\times\sqrt6}\)
cos θ = -6/6
cos θ = -1
θ = 180°
option (d) is correct.