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

`\hat{w}=\frac{\vec{5i+13j-k}}{13.92}`

Step-by-step 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* v`

`\vec{w}=(2i-j-3k)* (-3i+j-2k)`

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

`|w|=√(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}=\frac{\vec{w}}w`

`\hat{w}=\frac{\vec{5i+13j-k}}{13.92}`

Hence, this is the required solution.