Question

Find the distance between the point (7, 2, 4) and the plane determined by the points A (2, 5, -3), B (-2, -3, 5) and C (5, 3, -3).

Answer 1

Hint: We have been given three points. So firstly we will find the equation of plane by using the formula as follows:
Let us consider three points as $A\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right),B\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right),C\left( {{x}_{3}},{{y}_{3}},{{z}_{3}} \right)$.

Complete step-by-step answer:
Then, then equation of plane is $\left| \begin{matrix} x-{{x}_{1}} & y-{{y}_{1}} & z-{{z}_{1}} \\\ {{x}_{2}}-{{x}_{1}} & {{y}_{2}}-{{y}_{1}} & {{z}_{2}}-{{z}_{1}} \\\ {{x}_{3}}-{{x}_{1}} & {{y}_{3}}-{{y}_{1}} & {{z}_{3}}-{{z}_{1}} \\\ \end{matrix} \right|=0$
Also, we will use the formula of distance between a point $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ and the plane $Ax+By+Cz=D$ is as follows:
Distance $=\left| \dfrac{A{{x}_{1}}+B{{y}_{1}}+C{{z}_{1}}-D}{\sqrt{{{A}^{2}}+{{B}^{2}}+{{C}^{2}}}} \right|$.

We have been given three points $A\left( 2,5,-3 \right),B\left( -2,-3,5 \right),C\left( 5,3,-3 \right)$.
We know that the equation of a plane having point $A\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right),B\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right),C\left( {{x}_{3}},{{y}_{3}},{{z}_{3}} \right)$ is given as follows:

x-{{x}_{1}} & y-{{y}_{1}} & z-{{z}_{1}} \\\ {{x}_{2}}-{{x}_{1}} & {{y}_{2}}-{{y}_{1}} & {{z}_{2}}-{{z}_{1}} \\\ {{x}_{3}}-{{x}_{1}} & {{y}_{3}}-{{y}_{1}} & {{z}_{3}}-{{z}_{1}} \\\ \end{matrix} \right|=0$$ After using the above formula to find the equation of a plane, we get as follows: $$\begin{aligned} & \left| \begin{matrix} x-2 & y-5 & z-(-3) \\\ -2-2 & -3-5 & 5-(-3) \\\ 5-2 & 3-5 & -3-(-3) \\\ \end{matrix} \right|=0 \\\ & \left| \begin{matrix} x-2 & y-5 & z+3 \\\ -4 & -8 & 8 \\\ 3 & -2 & 0 \\\ \end{matrix} \right|=0 \\\ & \Rightarrow \left( x-2 \right)\left( -8\times 0-(-2)\times 8 \right)-(y-5)(-4\times 0-3\times 8)+(z+3)(-4\times -2(-8\times 3)=0 \\\ &\Rightarrow (x-2)(16)-(y-5)(-24)+(z+3)(8+24)=0 \\\ & \Rightarrow 16x-32+24y-120+32z+96=0 \\\ & \Rightarrow 16x+24y+32z-32-120+96=0 \\\ & \Rightarrow 16x+24y+32z-56=0 \\\ \end{aligned}$$ Taking 8 as common, we get as follows: $$\begin{aligned} & 8\left( 2x+3y+4z-7 \right)=0 \\\ & 2x+3y+4z-7=0 \\\ \end{aligned}$$ Hence the equation of the plane is $$2x+3y+4z-7=0$$. We know that the distance of a point $$\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$$ from plane $$ax+by+cz+d=0$$ is given as, $$d=\left| \dfrac{a{{x}_{1}}+b{{y}_{1}}+c{{z}_{1}}+d}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}} \right|$$ So we have been given the point (7, 2, 4). Equation of the plane is $$2x+3y+4z-7=0$$. $$d=\left| \dfrac{2\times 7+3\times 2+4\times 4-7}{\sqrt{{{2}^{2}}+{{3}^{2}}+{{4}^{2}}}} \right|=\left| \dfrac{14+6+16-7}{\sqrt{4+9+16}} \right|=\left| \dfrac{29}{\sqrt{29}} \right|$$ On rationalizing the above equation, we get as follows: $$d=\dfrac{29}{\sqrt{29}}\times \dfrac{\sqrt{29}}{\sqrt{29}}=\dfrac{29\times \sqrt{29}}{29}=\sqrt{29}$$ Therefore, the distance between the point and the plane is $$\sqrt{29}$$ units. Note: We can also use the formula $$a\left( x-{{x}_{1}} \right)+b\left( y-{{y}_{1}} \right)+c\left( z-{{z}_{1}} \right)=0$$ to find the equation of the plane and calculate the value of a, b and c by substituting other two points. Be careful while substituting values in the determinant and also while solving it as there is a chance of sign and calculation mistake.