121348
Find the angle between the two planes \(2 x+y-2 z=5\) and \(3 x-6 y-2 z=7\).
1 \(\cos ^{-1}(4 / 21)\)
2 \(\cos ^{-1}(2 / 21)\)
3 \(\cos ^{-1}(1 / 21)\)
4 \(\cos ^{-1}(5 / 21)\)
Explanation:
A The angle between two planes is the angle between their normals. From the equation of the planes, the normal vectors are \(\overrightarrow{\mathrm{n}}_1=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}\) and \(\overrightarrow{\mathrm{n}}_2=3 \hat{\mathrm{i}}-6 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}\) Therefore, \(\cos \theta=\left \vert\frac{\overrightarrow{\mathrm{n}}_1 \cdot \overrightarrow{\mathrm{n}}_2}{\left \vert\overrightarrow{\mathrm{n}_1}\right \vert\left \vert\overrightarrow{\mathrm{n}_2}\right \vert}\right \vert\)
\(=\left \vert\frac{(2 \hat{i}+\hat{j}-2 \hat{k}) \cdot(3 \hat{i}-6 \hat{j}-2 \hat{k})}{(\sqrt{4+1+4})(\sqrt{9+36+4})}\right \vert=\left(\frac{4}{21}\right)\)Hence, \(\theta=\cos ^{-1}\left(\frac{4}{21}\right)\)
BITSAT-2013
Three Dimensional Geometry
121392
The distance of the plane \(2 x-y-2 z-9=0\) from the origin is......units.
1 3
2 \(\sqrt{3}\)
3 1
4 9
Explanation:
A Given, Plane : \(2 \mathrm{x}-\mathrm{y}-2 \mathrm{z}-9=0\)
Distance of this plane from origin is -
\(\mathrm{d}=\frac{ \vert2 \times 0+(-1) \times 0+(-2) \times 0-9 \vert}{\sqrt{4+1+4}}=\frac{9}{\sqrt{9}}\)\(\Rightarrow \quad \mathrm{d}=3\)
121348
Find the angle between the two planes \(2 x+y-2 z=5\) and \(3 x-6 y-2 z=7\).
1 \(\cos ^{-1}(4 / 21)\)
2 \(\cos ^{-1}(2 / 21)\)
3 \(\cos ^{-1}(1 / 21)\)
4 \(\cos ^{-1}(5 / 21)\)
Explanation:
A The angle between two planes is the angle between their normals. From the equation of the planes, the normal vectors are \(\overrightarrow{\mathrm{n}}_1=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}\) and \(\overrightarrow{\mathrm{n}}_2=3 \hat{\mathrm{i}}-6 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}\) Therefore, \(\cos \theta=\left \vert\frac{\overrightarrow{\mathrm{n}}_1 \cdot \overrightarrow{\mathrm{n}}_2}{\left \vert\overrightarrow{\mathrm{n}_1}\right \vert\left \vert\overrightarrow{\mathrm{n}_2}\right \vert}\right \vert\)
\(=\left \vert\frac{(2 \hat{i}+\hat{j}-2 \hat{k}) \cdot(3 \hat{i}-6 \hat{j}-2 \hat{k})}{(\sqrt{4+1+4})(\sqrt{9+36+4})}\right \vert=\left(\frac{4}{21}\right)\)Hence, \(\theta=\cos ^{-1}\left(\frac{4}{21}\right)\)
BITSAT-2013
Three Dimensional Geometry
121392
The distance of the plane \(2 x-y-2 z-9=0\) from the origin is......units.
1 3
2 \(\sqrt{3}\)
3 1
4 9
Explanation:
A Given, Plane : \(2 \mathrm{x}-\mathrm{y}-2 \mathrm{z}-9=0\)
Distance of this plane from origin is -
\(\mathrm{d}=\frac{ \vert2 \times 0+(-1) \times 0+(-2) \times 0-9 \vert}{\sqrt{4+1+4}}=\frac{9}{\sqrt{9}}\)\(\Rightarrow \quad \mathrm{d}=3\)