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The image of point P(x, y, z) with respect to line x/1 = (y – 1)/2 = (z – 2)/3 is P′ (1, 0, 7). Find the coordinates of point P.
Direction ratios of a vector parallel to line (x–1)/2 = – y = (2z+1)/6 are :
Find the equation of the line which bisects the line segment joining points A(2, 3, 4) and B(4, 5, 8) and is perpendicular to the lines (x–8)/3 = (y+19)/(–16) = (z–10)/7 and (x–15)/3 = (y–29)/8 = (z–5)/(–5).
Find the equation of the line passing through the point of intersection of the lines x/1 = (y – 1)/2 = (z – 2)/3 and (x – 1)/0 = y/(– 3) = (z – 7)/2 and perpendicular to these given lines.
Two vertices of the parallelogram ABCD are given as A(– 1, 2, 1) and B(1, – 2, 5). If the equation of the line passing through C and D is (x – 4)/1 = (y + 7)/(– 2) = (z – 8)/2, then find the distance between sides AB and CD. Hence, find the area of parallelogram ABCD.
Find the distance between the line x/2 = (2y – 6)/4 = (1 – z)/(–1) and another line parallel to it passing through the point (4, 0, –5).
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