# Lecture - 11 Chapter 11 Three Dimensional Geometry

“Whenever you find yourself on the side of the majority, it is time to pause and reflect.” –Mark Twain

**1.** Find the distance of the point (2,5,-3) from the plane \(\vec{r}.(6\hat{i}-3\hat{j}+2\hat{k})=4\) .

**2.** In the following cases find the distance of each of the given points from the corresponding given plane: **a.** (3,-2,1) 3x-4y+12z=3 **b.** (2,3,-5) 2x-y+2z+3=0 **c.** (-6,0,0) 2x-3y+6z-2=0

**3.** Find the distance between the point P (6, 5, 9) and the plane determined by the pointsA(3,-1,2), B(5,2,4) and C(-1,-1,6).

**4.** Find the distance between the two planes: 2x+3y+4z=4 and 4x+6y+8z=12 .

**5.** Prove that if a plane has the intercepts a, b , c and is at the distance p units from the origin, then \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\frac{1}{p^2}\) .

**6.** Find the distance of the point whose position vector is \((2\hat{i}+\hat{j}-\hat{k})\) from the plane \(\vec{r}.(\hat{i}-2\hat{j}+4\hat{k})=9\) .

**7.** Find the distance between the planes \(\vec{r}.(\hat{i}+2\hat{j}+3\hat{k})+7=0\) & \(\vec{r}.(2\hat{i}+4\hat{j}+6\hat{k})+7=0\)

**8.** If the points (1,1,p) and (-3,0,1) be equidistant from the plane \(\vec{r}.(3\hat{i}+4\hat{j}-12\hat{k})+13=0\) , then find the value of p.