Let P be the foot of the perpendicular from the point Q(10,−3,−1) on the line 7x−3=−1y−2=−2z+2. Then the area of the right-angled triangle PQR, where R is the point (3,−2,1), is
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Step-by-Step Solution
Step 1: Find the coordinates of point P
First, we represent any point P on the given line in terms of a parameter λ. The direction vector of the line is d=(7,−1,−2). Since P is the foot of the perpendicular from Q to the line, the vector QP must be perpendicular to the direction vector of the line. We set their dot product to zero to find the value of λ, and then substitute λ back into the coordinates of P.
Step 2: Calculate the lengths of PQ and PR
Next, we calculate the lengths of the sides PQ and PR using the distance formula between two points. We express the coordinates of Q and R with a common denominator to simplify the calculations.
Step 3: Calculate the area of triangle PQR
The problem states that PQR is a right-angled triangle. Since P is the foot of the perpendicular from Q to the line, the angle ∠QPR is not necessarily 90∘. We check if the right angle is at Q by calculating the dot product of QP and QR. Since QP⋅QR=0, the right angle is at Q. Therefore, the area of the triangle is 21×PQ×QR. We calculate the length of QR and then the area.