The distance of the point P(3, 4, 4) from the point of intersection of the line joining the points. Q(3, $-$4, $-$5) and R(2, $-$3, 1) and the plane 2x + y + z = 7, is equal to ______________.
Answer (integer)
7
Solution
$$\overrightarrow {QR} : - {{x - 3} \over 1} = {{y + 4} \over { - 1}} = {{z + 5} \over { - 6}} = r$$<br><br>$\Rightarrow (x,y,z) \equiv (r + 3, - r - 4, - 6r - 5)$<br><br>Now, satisfying it in the given plane.<br><br>We get r = $-$2<br><br>so, required point of intersection is T(1, $-$2, 7).<br><br>Hence, PT = 7.
About this question
Subject: Mathematics · Chapter: Three Dimensional Geometry · Topic: Direction Cosines and Ratios
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