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: 7
Solutions · 1
Official solution
Direction of QR: R−Q=(−1,1,6). Line: (3−t,−4+t,−5+6t). Substitute into the plane:
\[2(3-t)+(-4+t)+(-5+6t)=7 \implies -3+5t=7 \implies t=2.\]
Point of intersection T=(1,−2,7).
\[PT=\sqrt{(3-1)^2+(4+2)^2+(4-7)^2}=\sqrt{4+36+9}=\sqrt{49}=7.\]