What is the unit vector that is normal to the plane containing 3i+7j-2k and 8i+2j+9k?

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What is the unit vector that is normal to the plane containing 3i+7j-2k and 8i+2j+9k?

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Answer 1 · 2016-11-07T03:13:58

The unit vector normal to the plane is
$(\frac{1}{94.01}) (67 ˆ i - 43 ˆ j + 50 ˆ k)$ $(1/94.01)(67hati-43hatj+50hatk)$ .

Explanation:

Let us consider $\to A = 3 ˆ i + 7 ˆ j - 2 ˆ k, \to B = 8 ˆ i + 2 ˆ j + 9 ˆ k$ $vecA=3hati+7hatj-2hatk, vecB= 8hati+2hatj+9hatk$
The normal to the plane $\to A, \to B$ $vecA, vecB$ is nothing but the vector perpendicular i.e., cross product of $\to A, \to B$ $vecA, vecB$ .
$\Rightarrow \to A \times \to B = ˆ i (63 + 4) - ˆ j (27 + 16) + ˆ k (6 - 56) = 67 ˆ i - 43 ˆ j + 50 ˆ k$ $=>vecAxxvecB= hati(63+4)-hatj(27+16)+hatk(6-56) =67hati-43hatj+50hatk$ .
The unit vector normal to the plane is
$\pm [\to A \times \to B / (∣ ∣ ∣ \to A \times \to B ∣ ∣ ∣)]$ $+-[vecAxxvecB//(|vecAxxvecB|)]$
So $∣ ∣ ∣ \to A \times \to B ∣ ∣ ∣ = \sqrt{{(67)}^{2} + {(- 43)}^{2} + {(50)}^{2}} = \sqrt{8838} = 94.01 \approx 94$ $|vecAxxvecB|=sqrt[(67)^2+(-43)^2+(50)^2]=sqrt8838=94.01~~94$
Now substitute all in above equation, we get unit vector = $\pm {[\frac{1}{\sqrt{8838}}] [67 ˆ i - 43 ˆ j + 50 ˆ k]}$ $+-{[1/(sqrt8838)][67hati-43hatj+50hatk]}$ .

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What is the unit vector that is normal to the plane containing 3i+7j-2k and 8i+2j+9k?

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