Practicing Success
A ball is thrown upwards from the plane surface of the ground. Suppose the plane surface from which the ball is thrown also 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 |
$\frac{5}{\sqrt{29}}$ units $\frac{7}{\sqrt{29}}$ units $\frac{6}{\sqrt{29}}$ units $\frac{8}{\sqrt{29}}$ units |
$\frac{5}{\sqrt{29}}$ units |
Maximum point D(2, 3, 1) so distance from place = $\frac{|3×2+2×3+4×1-11|}{\sqrt{3^2+2^2+4^2}} = \frac{|6+6+4-11|}{\sqrt{9+4+16}}$ $= \frac{5}{\sqrt{29}}$ units |