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Question Number 214838 by isaac_newton_2 last updated on 21/Dec/24

$$$$Determine the unit Vector perpendicular in plane of A = 2i-6j-3k , B = 4i+3j-k

Commented by TonyCWX08 last updated on 21/Dec/24

$$AreyousureAandBareplanes,andnotvectors?$$

Commented by mr W last updated on 21/Dec/24

$$hemeansunitvectorwhichis$$$$perpendiculartotheplanecontaining$$$$vectorsAandB.$$

Commented by TonyCWX08 last updated on 21/Dec/24

$$Ifthatso,thenalright.$$

Answered by mr W last updated on 21/Dec/24

$$C=A\times B=(2,-6,-3)\times (4,3,-1)$$$$=(15,-10,30)$$$$\frac{C}{\mid C\mid }=\frac{1}{\sqrt{{15}^2+{10}^2+{30}^2}}(15,-10,30)$$$$=(\frac{3}{7},-\frac{2}{7},\frac{6}{7})✓$$

Commented by mr W last updated on 21/Dec/24

Answered by MrGaster last updated on 21/Dec/24

$$\overrightarrow{A}=2\hat{i}-6\hat{j}-3\hat{k}$$$$\overrightarrow{B}=4\hat{i}+3\hat{j}-\hat{k}$$$$\overrightarrow{A}\times \overrightarrow{B}=\underset{\mathrm{D}eterminantexpansi\mathrm{on}}{\underbrace{\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& -6& -3\\ 4& 3& -1\end{array}\right|}}=\hat{i}((-6)\cdot (-1)-(-3)\cdot 3)-\hat{j}((2\cdot (-1)-(-3)\cdot 4)+\hat{k}(\left(2\right)\cdot 3-(-6)\cdot 4)$$$$=\hat{i}(6+9)-\hat{j}(-2+12)+\hat{k}(6+24)$$$$=15\hat{i}-10\hat{j}+30\hat{k}$$$$\mid \overrightarrow{A}\times \overrightarrow{B}\mid =\sqrt{{\left(15\right)}^2+{\left(10\right)}^2+{\left(30\right)}^2}$$$$=\sqrt{225+100+900}$$$$=\sqrt{1225}$$$$=35$$$$\hat{n}=\frac{\overrightarrow{A}\times \overrightarrow{B}}{\mid A\times B\mid }$$$$=\left(\frac{15}{35}\right)\hat{i}-\left(\frac{10}{35}\right)\hat{j}+\left(\frac{30}{35}\right)\hat{k}$$$$=\left(\frac{3}{7}\right)\hat{i}-\left(\frac{2}{7}\right)\hat{j}+\left(\frac{6}{7}\right)\hat{k}$$

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