Which of the following points lie on the plan | vedclass

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(C) Let the position vectors of the three points be $\vec{a} = 2 \hat{i}+3 \hat{j}-\hat{k}$,$\vec{b} = 3 \hat{i}+2 \hat{j}+\hat{k}$,and $\vec{c} = \ha

Vedclass · Accepted Answer

$2 \hat{i}-3 \hat{j}+13 \hat{k}$ and $2 \hat{i}+\frac{3}{2} \hat{j}+\frac{5}{2} \hat{k}$

Vedclass · Answer

(C) Let the position vectors of the three points be $\vec{a} = 2 \hat{i}+3 \hat{j}-\hat{k}$,$\vec{b} = 3 \hat{i}+2 \hat{j}+\hat{k}$,and $\vec{c} = \hat{i}+\hat{j}+4 \hat{k}$.The normal vector $\vec{n}$ to the plane is given by $\vec{n} = (\vec{b}-\vec{a}) 	imes (\vec{c}-\vec{a})$.$\vec{b}-\vec{a} = (3-2)\hat{i} + (2-3)\hat{j} + (1-(-1))\hat{k} = \hat{i}-\hat{j}+2\hat{k}$.$\vec{c}-\vec{a} = (1-2)\hat{i} + (1-3)\hat{j} + (4-(-1))\hat{k} = -\hat{i}-2\hat{j}+5\hat{k}$.$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -1 & 2 \ -1 & -2 & 5 \end{vmatrix} = \hat{i}(-5+4) - \hat{j}(5+2) + \hat{k}(-2-1) = -\hat{i}-7\hat{j}-3\hat{k}$.The equation of the plane is $(\vec{r}-\vec{a}) \cdot \vec{n} = 0$.$(x-2)(-1) + (y-3)(-7) + (z+1)(-3) = 0$.$-x+2 -7y+21 -3z-3 = 0 \Rightarrow -x-7y-3z+20 = 0 \Rightarrow x+7y+3z = 20$.Checking the points in option $C$:For $2 \hat{i}-3 \hat{j}+13 \hat{k}$: $2 + 7(-3) + 3(13) = 2 - 21 + 39 = 20$. (Satisfied)For $2 \hat{i}+\frac{3}{2} \hat{j}+\frac{5}{2} \hat{k}$: $2 + 7(\frac{3}{2}) + 3(\frac{5}{2}) = 2 + \frac{21}{2} + \frac{15}{2} = 2 + \frac{36}{2} = 2 + 18 = 20$. (Satisfied)Thus,the points in option $C$ lie on the plane.

Which of the following points lie on the plane passing through $2 \hat{i}+3 \hat{j}-\hat{k}$,$3 \hat{i}+2 \hat{j}+\hat{k}$ and $\hat{i}+\hat{j}+4 \hat{k}$?

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