Mathematics Questions and Answers – Three Dimensional Geometry – Angle between Two Planes – 1

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This set of Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Three Dimensional Geometry – Angle between Two Planes – 1”.

1. Which of the following is the correct formula for the angle between two planes?
a) cos⁡θ=\(\left |\frac{\vec{n_1}.\vec{n_2}}{|\vec{n_1}||\vec{n_2}|}\right |\)
b) sin⁡θ=\(\left |\frac{\vec{n_1}.\vec{n_2}}{|\vec{n_1}||\vec{n_2}|}\right |\)
c) cos⁡θ=\(\left |\frac{\vec{n_1}+\vec{n_2}}{|\vec{n_1}||\vec{n_2}|}\right |\)
d) sin⁡θ=\(\left |\frac{\vec{n_1}+\vec{n_2}}{(|\vec{n_1}|+|\vec{n_2}|}\right |\)
View Answer

Answer: a
Explanation: If two planes of the form \(\vec{r}.\vec{n_1}=d_1\) and \(\vec{r}.\vec{n_2}=d_2\) where \(\vec{n_1} \,and \,\vec{n_2}\) are the normals to the plane, then the angle between them is given by
cos⁡θ=\(\left |\frac{\vec{n_1}.\vec{n_2}}{|\vec{n_1}||\vec{n_2}|}\right |\)
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2. Which of the following is the correct formula for the angle between two planes \(A_1 x+B_1 y+C_1 z+D_1\)=0 and \(A_2 x+B_2 y+C_2 z+D_2\)=0?
a) cos⁡θ=\(\frac{A_1 B_1 C_1}{A_2 B_2 C_2}\)
b) cos⁡θ=\(\left |\frac{A_1 A_2+B_1 B_2+C_1 C_2}{\sqrt{A_1^2+B_1^2+C_1^2}\sqrt{A_2^2+B_2^2+C_2^2}}\right |\)
c) sin⁡θ=\(\left |\frac{A_1 A_2-B_1 B_2-C_1 C_2}{\sqrt{A_1^2+B_1^2+C_1^2}\sqrt{A_2^2+B_2^2+C_2^2}}\right |\)
d) cos⁡θ=\(A_1 A_2+B_1 B_2+C_1 C_2\)
View Answer

Answer: b
Explanation: If the planes are in the Cartesian form i.e. \(A_1 x+B_1 y+C_1 z+D_1\)=0 and \(A_2 x+B_2 y+C_2 z+D_2\)=0, where \(A_1,B_1,C_1 \,and \,A_2,B_2,C_2\) are the direction ratios of the planes, then the angle between them is given by
cos⁡θ=\(\left |\frac{A_1 A_2+B_1 B_2+C_1 C_2}{\sqrt{A_1^2+B_1^2+C_1^2} \sqrt{A_2^2+B_2^2+C_2^2}}\right |\)

3. Find the angle between two planes \(\vec{r}.(2\hat{i}-\hat{j}+\hat{k})=3\) and \(\vec{r}.(3\hat{i}+2\hat{j}-3\hat{k})\)=5.
a) \(cos^{-1}⁡\frac{1}{\sqrt{22}}\)
b) \(cos^{-1}⁡\frac{1}{\sqrt{6}}\)
c) \(cos^{-1}⁡\frac{1}{\sqrt{132}}\)
d) \(cos^{-1}⁡\frac{1}{\sqrt{13}}\)
View Answer

Answer: c
Explanation: Given that, the normal to the planes are \(\vec{n_1}=2\hat{i}-\hat{j}+\hat{k}\) and \(\vec{n_2}=3\hat{i}+2\hat{j}-3\hat{k}\)
cos⁡θ=\(\left |\frac{\vec{n_1}.\vec{n_2}}{|\vec{n_1}||\vec{n_2}|}\right |\)
\(|\vec{n_1}|=\sqrt{2^2+(-1)^2+1^2}=\sqrt{6}\)
\(|\vec{n_2}|=\sqrt{3^2+2^2+(-3)^2}=\sqrt{22}\)
\(\vec{n_1}.\vec{n_2}\)=2(3)-1(2)+1(-3)=6-2-3=1
cos⁡θ=\(\frac{1}{\sqrt{22}.\sqrt{6}}=\frac{1}{\sqrt{132}}\)
∴θ=\(cos^{-1}⁡\frac{1}{\sqrt{132}}\).
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4. If the planes \(A_1 x+B_1 y+C_1 z+D_1\)=0 and \(A_2 x+B_2 y+C_2 z+D_2\)=0 are at right angles to each other, then which of the following is true?
a) \(\frac{A_1+B_1+C_1}{A_2+B_2+C_2}\)=0
b) \(A_1+A_2+B_1 +B_2+C_1+C_2\)=0
c) \(A_1+B_1+C_1=A_2 B_2 C_2\)
d) \(A_1 A_2+B_1 B_2+C_1 C_2\)=0
View Answer

Answer: d
Explanation: We know that the angle between two planes is given by
cos⁡θ=\(\left |\frac{A_1 A_2+B_1 B_2+C_1 C_2}{\sqrt{A_1^2+B_1^2+C_1^2} \sqrt{A_2^2+B_2^2+C_2^2}}\right |\)
Given that, θ=90°
∴cos⁡90°=\(\left |\frac{A_1 A_2+B_1 B_2+C_1 C_2}{\sqrt{A_1^2+B_1^2+C_1^2}\sqrt{A_2^2+B_2^2+C_2^2}}\right |\)
0=\(\left |\frac{A_1 A_2+B_1 B_2+C_1 C_2}{\sqrt{A_1^2+B_1^2+C_1^2}\sqrt{A_2^2+B_2^2+C_2^2}}\right |\)
⇒\(A_1 A_2+B_1 B_2+C_1 C_2\)=0.

5. Find the angle between the planes 6x-3y+7z=8 and 2x+3y-2z=5?
a) \(cos^{-1}\frac{⁡11}{\sqrt{98}}\)
b) \(cos^{-1}⁡\frac{⁡11}{\sqrt{1598}}\)
c) \(cos^{-1}\frac{⁡⁡13}{\sqrt{198}}\)
d) \(cos^{-1}\frac{⁡⁡11}{1598}\)
View Answer

Answer: b
Explanation: We know that, the angle between two planes of the form \(A_1 x+B_1 y+C_1 z+D_1\)=0 and \(A_2 x+B_2 y+C_2 z+D_2\)=0 is given by
cos⁡θ=\(\left |\frac{A_1 A_2+B_1 B_2+C_1 C_2}{\sqrt{A_1^2+B_1^2+C_1^2} \sqrt{A_2^2+B_2^2+C_2^2}}\right |\)
Given that, \(A_1=6,B_1=-3,C_1=7\) and \(A_2=2,B_2=3,C_2=-2\)
cos⁡θ=\(\left |\frac{6(2)-3(3)+7(-2)}{|\sqrt{6^2+(-3)^2+7^2} \sqrt{2^2+3^2+(-2)^2}|}\right |\)
cos⁡θ=\(|\frac{-11}{\sqrt{94}.\sqrt{17}}|\)
θ=\(cos^{-1}⁡\frac{⁡11}{\sqrt{1598}}\).
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6. If two vectors \(\vec{r}.\vec{n_1}=d_1\) and \(\vec{r}.\vec{n_2}=d_2\) are such that \(\vec{n_1}.\vec{n_2}\)=0, then which of the following is true?
a) The planes are perpendicular to each other
b) The planes are parallel to each other
c) Depends on the value of the vector
d) The planes are at an angle greater than 90°
View Answer

Answer: a
Explanation: We know that if the scalar or dot product of vectors is 0, then they are at right angles to each other. Here the dot product of the normal vectors of the plane is 0, i.e. \(\vec{n_1}.\vec{n_2}\)=0. Hence, the planes will be perpendicular to each other.

7. Find the angle between the two planes 2x+2y+z=2 and x-y+z=1?
a) \(cos^{-1}\frac{⁡1}{3}\)
b) \(cos^{-1}⁡\sqrt{3}\)
c) \(cos^{-1}⁡\frac{1}{3}\)
d) \(cos^{-1}⁡\frac{1}{3\sqrt{3}}\)
View Answer

Answer: d
Explanation: The angle between two planes of the form \(A_1 x+B_1 y+C_1 z+D_1\)=0 and \(A_2 x+B_2 y+C_2 z+D_2\)=0 is given by
cos⁡θ=\(\left |\frac{A_1 A_2+B_1 B_2+C_1 C_2}{\sqrt{A_1^2+B_1^2+C_1^2} \sqrt{A_2^2+B_2^2+C_2^2}}\right |\)
According to the given question, \(A_1=2,B_1=2,C_1=1 \,and \,A_2=1,B_2=-1,C_2=1\)
cos⁡θ=\(\left |\frac{2(1)+2(-1)+1(1)}{|\sqrt{2^2+2^2+1^2} \sqrt{1^2+(-1)^2+1^2}|}\right |\)
cos⁡θ=\(\left |\frac{1}{\sqrt{9}.\sqrt{3}}\right |\)
θ=\(cos^{-1}⁡\frac{1}{3\sqrt{3}}\).
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8. Which of the given set of planes are perpendicular to each other?
a) \(\vec{r}.(2\hat{i}+2\hat{j}+\hat{k})\)=5 and \(\vec{r}.(\hat{i}+2\hat{j}+2\hat{k})\)=5
b) \(\vec{r}.(\hat{i}-2\hat{j}+\hat{k})\)=7 and \(\vec{r}.(\hat{i}+\hat{j}+2\hat{k})\)=2
c) \(\vec{r}.(2\hat{i}-2\hat{j}+\hat{k})\)=4 and \(\vec{r}.(\hat{i}+2\hat{j}+2\hat{k})\)=5
d) \(\vec{r}.(3\hat{i}-2\hat{j}+\hat{k})\)=2 and \(\vec{r}.(\hat{i}+2\hat{j}+8\hat{k})\)=8
View Answer

Answer: c
Explanation: Consider the set of planes \(\vec{r}.(2\hat{i}-2\hat{j}+\hat{k})=4 \,and \,\vec{r}.(\hat{i}+2\hat{j}+2\hat{k})\)=5
For the set of planes to be perpendicular \(\vec{n_1.}\vec{n_2}\)=0
In the above set of planes, \(\vec{n_1}=2\hat{i}-2\hat{j}+\hat{k}\) and \(\vec{n_2}=\hat{i}+2\hat{j}+2\hat{k}\)
∴\(\vec{n_1}.\vec{n_2}\)=2(1)-2(2)+1(2)=0
Hence, they are perpendicular.

9. Find the angle between the planes \(\vec{r}.(4\hat{i}+\hat{j}-2\hat{k})\)=6 and \(\vec{r}.(5\hat{i}-6\hat{j}+\hat{k})\)=7?
a) \(cos^{-1}⁡\frac{12}{\sqrt{1302}}\)
b) \(cos^{-1}⁡\frac{1}{\sqrt{1392}}\)
c) \(cos^{-1}\frac{⁡23}{\sqrt{102}}\)
d) \(cos^{-1}⁡\frac{15}{\sqrt{134}}\)
View Answer

Answer: a
Explanation: Given that, the normal to the planes are \(\vec{n_1}=4\hat{i}+\hat{j}-2\hat{k} \,and \,\vec{n_2}=5\hat{i}-6\hat{j}+\hat{k}\)
The angle between two planes of the form \(\vec{r}.\vec{n_1}=d_1 \,and \,\vec{r}.\vec{n_2}=d_2\) is given by
cos⁡θ=\(\left |\frac{\vec{n_1}.\vec{n_2}}{|\vec{n_1}||\vec{n_2}|}\right |\)
\(|\vec{n_1}|=\sqrt{4^2+1^2+(-2)^2}=\sqrt{21}\)
\(|\vec{n_2}|=\sqrt{5^2+(-6)^2+1^2}=\sqrt{62}\)
\(\vec{n_1}.\vec{n_2}\)=4(5)+1(-6)-2(1)=20-6-2=12
cos⁡θ=\(\frac{12}{\sqrt{21}.\sqrt{62}}=\frac{12}{\sqrt{1302}}\) ∴θ=\(cos^{-1}⁡\frac{12}{\sqrt{1302}}\).
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10. Which of the following sets of planes are parallel to each other?
a) 2x+3y+4z=8 and 3x+9y+12z=7
b) 2x+3y+4z=2 and 4x+6y+8z=9
c) 3x+2y+4z=0 and 3x+4y+2z=0
d) 2x+4y+8z=9 and 4x+2y+7z=0
View Answer

Answer: b
Explanation: If two planes are of the form \(A_1 x+B_1 y+C_1 z+D_1\)=0 and \(A_2 x+B_2 y+C_2 z+D_2\)=0, then they will be parallel if \(\frac{A_1}{A_2}=\frac{B_1}{B_2}=\frac{C_1}{C_2}\)
Consider the planes 2x+3y+4z=2 and 4x+6y+8z=9
\(\frac{A_1}{A_2}=\frac{2}{4}\)
\(\frac{B_1}{B_2}=\frac{3}{6}\)
\(\frac{C_1}{C_2}=\frac{4}{8}\)
\(\frac{2}{4}=\frac{3}{6}=\frac{4}{8}=\frac{1}{2}\)
Hence, the above set of planes are parallel.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter