Question

5.Let A be an arbitrary vector and let n be a unit vector in some fixed direction. Show that A=(A*widehat n)widehat n+(widehat n times A)times widehat n ..

Answer 1

Explanation & proof

• The vector equation given in the description is

$\qquad\qquad \small A=(A*\widehat n)\widehat n+(\widehat n \times A) \times \widehat n$

• The first term is some product of the three vectors where as the second is a vector product of the three.
• In general, if a, b & c are three non-coplanar vectors then,

$\qquad\qquad \begin{aligned} \small \widehat a\times (\widehat b\times \widehat c)&= \small (\widehat a.\widehat c)\widehat b-(\widehat a.\widehat b)\widehat c \end{aligned}$ & $\qquad\qquad \begin{aligned} \small \widehat a\times (\widehat b\times \widehat c) = -(\widehat b\times \widehat c)\times \widehat a \end{aligned}$

and $\small \widehat a.\widehat b=\widehat b.\widehat a$

• Therefore, by applying these to the given one

$\qquad\qquad \begin{aligned} \small \widehat A &= \small (\widehat A*\widehat n)\widehat n+(\widehat n \times \widehat A)\times \widehat n\\ &= \small (\widehat A.\widehat n)\widehat n-\widehat n\times (\widehat n \times \widehat A)\\ &= \small (\widehat A.\widehat n)\widehat n-\big[(\widehat n.\widehat A)\widehat n-(\widehat n.\widehat n)\widehat A \big]\\ &= \small \cancel{(\widehat A.\widehat n)\widehat n}-\cancel{(\widehat A.\widehat n)\widehat n}+(\widehat n.\widehat n)\widehat A\\ &= \small (1)\widehat A\\ &= \small \widehat A \end{aligned}$