If a→,b→,c→are three unit vectors, such that a→.b→=0=a→.c→ and the angle between b→ and c→ is π/3 then the value of |a→×b→-a→×c→|is

Solution:

$\begin{array}{l} {| \vec{a} \times \vec{b} - \vec{a} \times \vec{c} |}^{2} \\ = {| \vec{a} \times (\vec{b} - \vec{c}) |}^{2} \\ = {| \vec{a} |}^{2} {| \vec{b} - \vec{c} |}^{2} - {(\vec{a} . (\vec{b} - \vec{c}))}^{2} = {| \vec{b} - \vec{c} |}^{2} \\ = {| \vec{b} |}^{2} + {| \vec{c} |}^{2} - 2 | \vec{b} | | \vec{c} | cos \frac{π}{3} = 1 \end{array}$

If $\vec{a}, \vec{b}, \vec{c}$ are three unit vectors, such that $\vec{a} . \vec{b} = 0 = \vec{a} . \vec{c}$ and the angle between $\vec{b} a n d \vec{c} is π / 3$ then the value of $| \vec{a} \times \vec{b} - \vec{a} \times \vec{c} |$ is

Solution:

Related content