Question

Find a unit vector normal to the plane containing Bold u equals 2 Bold i minus Bold j minus 3 Bold k and Bold v equals negative 3 Bold i plus Bold j minus 2 Bold k.

Answer 1

Answer:

$\hat{w}=\dfrac{\vec{5i+13j-k}}{13.92}$

Explanation:

It is given that,

$\vec{u}=2i-j-3k$

$\vec{v}=-3i+j-2k$

Taking the cross product of v and v such that,

$\vec{w}=u\times v$

$\vec{w}=(2i-j-3k)\times (-3i+j-2k)$

$\vec{w}=5i+13j-k$

$|w|=\sqrt{5^2+13^2(-1)^2}$

|w| = 13.92

Let $\hat{w}$ is the unit vector normal to the plane containing u and v. So,

$\hat{w}=\dfrac{\vec{w}}{|w|}$

$\hat{w}=\dfrac{\vec{5i+13j-k}}{13.92}$

Hence, this is the required solution.