Aspire Faculty ID #17314 · Topic: CUET UG 2022 Applied Mathematics · Just now
CUET UG 2022 Applied Mathematics

A ball is thrown upwards from the plane surface of the ground.  
Suppose the plane surface from which the ball is thrown consists of the points  

$A(1,0,2),\; B(3,-1,1)$ and $C(1,2,1)$ on it.  

The highest point of the ball takes is $D(2,3,1)$ as shown in the figure.

Using this information answer the question.
The maximum height of the ball from the ground is:

Solution

Equation of plane through $A,B,C$

$3x+2y+4z=11$

Maximum height of the ball = perpendicular distance of point $D(2,3,1)$ from the plane.

Distance formula

$d=\frac{|Ax_1+By_1+Cz_1+D|}{\sqrt{A^2+B^2+C^2}}$

Plane

$3x+2y+4z-11=0$

Substitute $D(2,3,1)$

$=|3(2)+2(3)+4(1)-11|$

$=|6+6+4-11|$

$=|5|$

Denominator

$\sqrt{3^2+2^2+4^2}$

$=\sqrt{9+4+16}$

$=\sqrt{29}$


Therefore

$d=\frac{5}{\sqrt{29}}$
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