Shortest distance between lines \(\frac{(x-5)}{4}\)=\(\frac{(y-3)}{6}\)=\(\frac{(z-2)}{4}\) and \(\frac{(x-3)}{7}=\frac{(y-2)}{5}=\frac{(z-9)}{6}\) is ?
A rectangular parallelepiped with edges along x, y, z axis has length 3, 4, 5 respectively. Fiind the shortest distance of the body diagonal from one of the edges parallel to z-axis which is skew to the diagonal
Let Q and R be two points on the line \(\frac{x+1}{2}=\frac{y+2}{3}=\frac{z−1}{2}\) at a distance \(\sqrt{26}\) from the point P(4, 2, 7). Then the square of the area of the triangle PQR is _______ .
Let the plane\(P : \stackrel{→}{r} . \stackrel{→}{a} = d\)contain the line of intersection of two planes\(\stackrel{→}{r} . ( \hat{i} + 3\hat{j} - \hat{k} ) = 6\)and\(\stackrel{→}{r} . ( -6\hat{i} + 5\hat{j} - \hat{k} ) = 7\). If the plane P passes through the point (2, 3, 1/2), then the value of \(\frac{| 13a→|² }{d²}\) is equal to
Let Q be the mirror image of the point P(1, 2, 1) with respect to the plane x + 2y + 2z = 16. Let T be a plane passing through the point Q and contains the line\(\vec{r}=−\hat{k}+λ(\hat{i}+\hat{j}+2\hat{k}), λ ∈ R.\)Then, which of the following points lies on T?