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Amanda [17]
3 years ago
5

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.
Physics
1 answer:
Nastasia [14]3 years ago
7 0

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.

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