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Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
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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