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Question Number 143127 by liberty last updated on 10/Jun/21

Answered by EDWIN88 last updated on 11/Jun/21

$$\mathrm{Let}\{\begin{array}{l}\overrightarrow{\mathrm{a}}=\mathrm{QR}=(1,3,-1)\\ \overrightarrow{\mathrm{b}}=\mathrm{QS}=(-3,2,3)\\ \overrightarrow{\mathrm{c}}=\mathrm{QP}=(0,3,3)\end{array}$$$$\mathrm{distance}\mathrm{d}\mathrm{from}\mathrm{P}\mathrm{to}\mathrm{plane}\mathrm{is}$$$$\mathrm{d}=\frac{\mid \overrightarrow{\mathrm{a}}.\left(\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{c}}\right)\mid }{\mid \overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}\mid }$$$$\left(1\right)\overrightarrow{\mathrm{a}}.\left(\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{c}}\right)=\left|\begin{array}{c}13-1\\ -323\\ 033\end{array}\right|$$$$=1(-3)-3(-9)-1(-9)$$$$=-3+27+9=33$$$$\left(2\right)\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}=\left|\begin{array}{c}13-1\\ -323\end{array}\right|$$$$=(11,0,11)$$$$\Rightarrow \mathrm{distance}=\frac{33}{\sqrt{121+121}}=\frac{33}{11\sqrt{2}}=\frac{3}{\sqrt{2}}$$$$\Rightarrow \mathrm{distance}=\frac{3}{\sqrt{2}}◻$$

Commented by EDWIN88 last updated on 11/Jun/21

$$\mathrm{wrong}.$$$$\mathrm{it}\mathrm{should}\mathrm{be}\mathrm{d}=\frac{\mid \overrightarrow{\mathrm{b}}.\left(\overrightarrow{\mathrm{c}}\times \overrightarrow{\mathrm{a}}\right)\mid }{\mid \overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{c}}\mid }$$$$\mathrm{where}\overrightarrow{\mathrm{b}}.\left(\overrightarrow{\mathrm{c}}\times \overrightarrow{\mathrm{a}}\right)=\left|\begin{array}{c}13-1\\ -323\\ 033\end{array}\right|$$$$=1(-3)-3(-9)-1(-9)$$$$=-3+27+9=33$$$$\mathrm{and}\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{c}}=\left|\begin{array}{c}13-1\\ -323\end{array}\right|$$$$=(11,0,11)$$$$\mathrm{therefore}\mathrm{d}=\frac{33}{\sqrt{121+121}}=\frac{33}{11\sqrt{2}}=\frac{3}{\sqrt{2}}$$

Commented by liberty last updated on 10/Jun/21

$$If\{\begin{array}{l}\overrightarrow{a}=(0,3,3)\\ \overrightarrow{b}=(1,3,-1)\\ \overrightarrow{c}=(-3,2,3)\end{array}$$$$\Rightarrow d=\frac{\mid \overrightarrow{a}.\left(\overrightarrow{b}\times \overrightarrow{c}\right)\mid }{\mid \overrightarrow{a}\times \overrightarrow{b}\mid }$$$$\overrightarrow{a}.\left(\overrightarrow{b}\times \overrightarrow{c}\right)=\left|\begin{array}{c}033\\ 13-1\\ -323\end{array}\right|$$$$=(11,0,11)$$$$\overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{c}033\\ 13-1\end{array}\right|$$$$=(-12,3,-3)$$$$\Rightarrow d=\frac{11\sqrt{2}}{\sqrt{144+9+9}}=\frac{11\sqrt{2}}{9\sqrt{2}}=\frac{11}{9}?$$

Answered by mr W last updated on 11/Jun/21

$$\mathit{a}=\overrightarrow{QR}=(1,3,-1)$$$$\mathit{b}=\overrightarrow{QS}=(-3,2,3)$$$$\mathit{c}=\overrightarrow{QP}=(0,3,3)$$$$\mathit{a}\times \mathit{b}=\left|\begin{array}{ccc}1& 3& -1\\ -3& 2& 3\end{array}\right|=(11,0,11)$$$$d=\frac{\mathit{c}\cdot \left(\mathit{a}\times \mathit{b}\right)}{\mid \mathit{a}\times \mathit{b}\mid }=\frac{0\times 11+3\times 0+3\times 11}{\sqrt{{11}^2+0^2+{11}^2}}=\frac{3}{\sqrt{2}}$$

Commented by mr W last updated on 10/Jun/21

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