A line L₁ passing through the point A with position vector a =4 +2 +2 k is parallel to the vector b =2 +3 +6 k . The length of the perpendicular drawn from a point P with position vector p = +2 +3 k to L₁ is

Step-by-step solution

The line L₁ passes through point A with position vector a = 4 i + 2 j + 2 k and is parallel to the vector b = 2 i + 3 j + 6 k . The position vector of point P is p = i + 2 j + 3 k . The vector AP = p - a = (1-4) i + (2-2) j + (3-2) k = -3 i + 0 j + k . The length of the perpendicular from P to the line L₁ is given by the formula d = | AP b | | b | . First, calculate the cross product AP b : AP b = vmatrix i & j & k -3 & 0 & 1 2 & 3 & 6 vmatrix = i (0 - 3) - j (-18 - 2) + k (-9 - 0) = -3 i + 20 j - 9 k . Calculate t