Lecture 11 Three Dimensional Geometry

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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.

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