Mathematics Questions and Answers – Three Dimensional Geometry – Plane

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This set of Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Three Dimensional Geometry – Plane”.

1. Which of the following is not the correct formula for representing a plane?
a) \(\vec{r}.\hat{n}=d\)
b) ax+by+cz=d
c) lx+my+nz=d
d) al+mb+cn=d2
View Answer

Answer: d
Explanation: \(\vec{r}.\hat{n}=d\), ax+by+cz=d, lx+my+nz=d are the various ways of representing a plane.
\(\vec{r}.\hat{n}\)=d is the vector equation of the plane, where \(\hat{n}\) is the unit vector normal to the plane.
ax+by+cz=d, lx+my+nz=d are the Cartesian equation of the plane in the normal form where, a, b, c are the direction ratios and l, m, n are the direction cosines of the normal to the plane respectively.
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2. If the plane passes through three collinear points \((x_1,y_1,z_1),(x_2,y_2,z_2),(x_3,y_3,z_3)\) then which of the following is true?
a) \(x_1 y_1 z_1+x_2 y_2 z_2+x_3 y_3 z_3\)=0
b) \(\begin{vmatrix}x_1&y_1&z_1\\x_2&y_2&z_2\\x_3&y_3&z_3\end{vmatrix}\)=0
c) \(\begin{vmatrix}x_1\\y_2\\z_3\end{vmatrix}\)=0
d) \(x_1 x_2 x_3+y_1 y_2 y_3+z_1 z_2 z_3=0\)
View Answer

Answer: b
Explanation: If the three points are collinear, then they will be in a straight line and hence the determinant of the three points will be zero.
i.e.\(\begin{vmatrix}x_1&y_1&z_1\\x_2&y_2&z_2\\x_3&y_3&z_3\end{vmatrix}\)=0

3. Find the vector equation of the plane which is at a distance of \(\frac{7}{\sqrt{38}}\) from the origin and the normal vector from origin is \(2\hat{i}+3\hat{j}-5\hat{k}\)?
a) \(\vec{r}.(\frac{2\hat{i}}{38}+\frac{3\hat{j}}{\sqrt{38}}-\frac{5\hat{k}}{\sqrt{38}})=\frac{7}{\sqrt{56}}\)
b) \(\vec{r}.(\frac{2\hat{i}}{\sqrt{38}}+\frac{3\hat{j}}{\sqrt{38}}-\frac{5\hat{k}}{\sqrt{38}})=\frac{7}{\sqrt{38}}\)
c) \(\vec{r}.(\frac{2\hat{i}}{\sqrt{38}}+\frac{5\hat{j}}{\sqrt{38}}+\frac{3\hat{k}}{\sqrt{38}})=\frac{7}{\sqrt{38}}\)
d) \(\vec{r}.(\frac{2\hat{i}}{\sqrt{58}}-\frac{3\hat{j}}{\sqrt{37}}-\frac{5\hat{k}}{\sqrt{38}})=\frac{7}{\sqrt{38}}\)
View Answer

Answer: b
Explanation: Let \(\vec{n}=2\hat{i}+3\hat{j}-5\hat{k}\)
\(\hat{n}=\frac{\vec{n}}{|\vec{n}|}=\frac{(2\hat{i}+3\hat{j}-5\hat{k})}{\sqrt{(2^2+3^2+(-5)^2)}}=\frac{(2\hat{i}+3\hat{j}-5\hat{k})}{\sqrt{38}}\)
Hence, the required equation of the plane is \(\vec{r}.(\frac{2\hat{i}}{\sqrt{38}}+\frac{3\hat{j}}{\sqrt{38}}-\frac{5\hat{k}}{\sqrt{38}})=\frac{7}{\sqrt{38}}\)
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4. Find the equation of the plane passing through three points (1,2,-1), (0,-1,2) and (3,1,1).
a) \((\vec{r}-(\hat{i}+2\hat{j}-\hat{k})).[(-\hat{j}+2\hat{k})×(2\hat{i}-\hat{j}+2\hat{k})]\)=0
b) \((\vec{r}-(\hat{i}+2\hat{j}-\hat{k})).[(\hat{i}-3\hat{j}+3\hat{k})×(2\hat{i}-\hat{j}+2\hat{k})]\)=0
c) \((\vec{r}-(\hat{i}+2\hat{j}-\hat{k})).[(3\hat{i}+\hat{j}+\hat{k})×(2\hat{i}-\hat{j}+2\hat{k})]\)=0
d) \((\vec{r}-(\hat{i}+2\hat{j})).[(-\hat{i}-3\hat{j}+3\hat{k})×(2\hat{i}-\hat{j}+2\hat{k})]\)=0
View Answer

Answer: b
Explanation: Let \(\vec{a}=\hat{i}+2\hat{j}-\hat{k}, \,\vec{b}=-\hat{j}+2\hat{k}, \,\vec{c}=3\hat{i}+\hat{j}+\hat{k}\)
The vector equation of the plane passing through three points is given by
\((\vec{r}-\vec{a}).[(\vec{b}-\vec{a})×(\vec{c}-\vec{a})]\)=0
\((\vec{r}-(\hat{i}+2\hat{j}-\hat{k})).[((-\hat{j}+2\hat{k})-(\hat{i}+2\hat{j}-\hat{k}))×((3\hat{i}+\hat{j}+\hat{k})–(\hat{i}+2\hat{j}-\hat{k}))]\)=0
\((\vec{r}-(\hat{i}+2\hat{j}-\hat{k})).[(\hat{i}-3\hat{j}+3\hat{k})×(2\hat{i}-\hat{j}+2\hat{k})]\)=0

5. Find the Cartesian equation of the plane \(\vec{r}.(2\hat{i}+\hat{j}-\hat{k})\)=4.
a) x+y-z=-4
b) 2x+y-z=4
c) x+y+z=4
d) -2x-y+z=4
View Answer

Answer: b
Explanation: Given that the equation of the plane is \(\vec{r}.(2\hat{i}+\hat{j}-\hat{k})\)=4
We know that, \(\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}\)
∴\((x\hat{i}+y\hat{j}+z\hat{k}).(2\hat{i}+\hat{j}-\hat{k})\)=4
⇒2x+y-z=4 is the Cartesian equation of the plane.
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6. Find the vector equation of the plane passing through the point (2,1,-1) and normal to the plane is \(2\hat{i}+\hat{j}-3\hat{k}\)?
a) \((\vec{r}-(2\hat{i}-7\hat{k})).(2\hat{i}+\hat{j}-3\hat{k})\)=0
b) \((\vec{r}-(2\hat{i}+3\hat{j}-\hat{k})).(2\hat{i}-3\hat{k})\)=0
c) \((\vec{r}-(\hat{i}+\hat{j}-3\hat{k})).(2\hat{i}+6\hat{j}-3\hat{k})\)=0
d) \((\vec{r}-(2\hat{i}+\hat{j}-\hat{k})).(2\hat{i}+\hat{j}-3\hat{k})\)=0
View Answer

Answer: d
Explanation: The position vector of the point (2,1,-1) is \(\vec{a}=2\hat{i}+\hat{j}-\hat{k}\) and the normal vector \(\vec{N}\) perpendicular to the plane is \(\vec{N}=2\hat{i}+\hat{j}-3\hat{k}\)
The vector equation of the plane is given by \((\vec{r}-\vec{a}).\vec{N}\)=0
Therefore, \((\vec{r}-(2\hat{i}+\hat{j}-\hat{k})).(2\hat{i}+\hat{j}-3\hat{k})\)=0

7. Find the distance of the plane 3x+4y-5z-7=0.
a) \(\frac{7}{\sqrt{40}}\)
b) \(\frac{6}{\sqrt{34}}\)
c) \(\frac{8}{\sqrt{50}}\)
d) \(\frac{7}{\sqrt{50}}\)
View Answer

Answer: d
Explanation: From the given equation, the direction ratios of the normal to the plane are 3, 4, -5; the direction cosines are
\(\frac{3}{\sqrt{3^2+4^2+(-5)^2}},\frac{4}{\sqrt{3^2+4^2+(-5)^2}},\frac{-5}{\sqrt{3^2+4^2+(-5)^2}}\),i.e. \(\frac{3}{\sqrt{50}},\frac{4}{\sqrt{50}},\frac{-5}{\sqrt{50}}\)
Dividing the equation throughout by √50, we get
\(\frac{3}{\sqrt{50}} x+\frac{4}{\sqrt{50}} y-\frac{5}{\sqrt{50}} z=\frac{7}{\sqrt{50}}\)
The above equation is in the form of lx+my+nz=d, where d is the distance of the plane from the origin. So, the distance of the plane from the origin is \(\frac{7}{\sqrt{50}}\).
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8. Find the Cartesian equation of the plane \(\vec{r}.[(λ+2μ) \hat{i}+(2λ-μ) \hat{j}+(3λ-2μ)\hat{k}]\)=12.
a) (λ-μ)x+y+(3λ-2μ)z=12
b) (λ+3μ)x+(2+μ)y+(3λ-2μ)z=12
c) (λ+2μ)x-2λy+(3λ-2μ)z=12
d) (λ+2μ)x+(2λ-μ)y+(3λ-2μ)z=12
View Answer

Answer: d
Explanation: Given that the equation of the plane is \(\vec{r}.[(λ+2μ) \hat{i}+(2λ-μ) \hat{j}+(3λ-2μ)\hat{k}]\)=12
We know that, \(\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}\)
∴\((x\hat{i}+y\hat{j}+z\hat{k}).([(λ+2μ) \hat{i}+(2λ-μ) \hat{j}+(3λ-2μ)\hat{k}])\)=12
⇒(λ+2μ)x+(2λ-μ)y+(3λ-2μ)z=12 is the Cartesian equation of the plane.

9. Find the equation of the plane passing through the three points (2,2,0), (1,2,1), (-1,2,-2).
a) \((\vec{r}-(2\hat{i}+2\hat{j})).[(-\hat{i}+\hat{k})×(-3\hat{i}-2\hat{k})]\)=0
b) \((\vec{r}-(3\hat{i}-2\hat{k})).[(-\hat{i}+\hat{k})×(2\hat{i}-2\hat{j})]\)=0
c) \((\vec{r}+(2\hat{i}+2\hat{j})).[(-\hat{i}-\hat{k})×(-3\hat{i}-2\hat{k})]\)=0
d) \((\vec{r}-(2\hat{i}+2\hat{j})).[(-\hat{i}-\hat{k})×(3\hat{i}+2\hat{k})]\)=0
View Answer

Answer: a
Explanation: Let \(\vec{a}=2\hat{i}+2\hat{j}, \,\vec{b}=\hat{i}+2\hat{j}+\hat{k}, \,\vec{c}=-\hat{i}+2\hat{j}-2\hat{k}\)
The vector equation of the plane passing through three points is given by
\((\vec{r}-\vec{a}).[(\vec{b}-\vec{a})×(\vec{c}-\vec{a})]\)=0
\((\vec{r}-(2\hat{i}+2\hat{j})).[((\hat{i}+2\hat{j}+\hat{k})-(2\hat{i}+2\hat{j}))×((-\hat{i}+2\hat{j}-2\hat{k})–(2\hat{i}+2\hat{j}))]\)=0
\((\vec{r}-(2\hat{i}+2\hat{j})).[(-\hat{i}+\hat{k})×(-3\hat{i}-2\hat{k})]\)=0.
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10. Find the Cartesian equation of the plane passing through the point (3,2,-3) and the normal to the plane is \(4\hat{i}-2\hat{j}+5\hat{k}\)?
a) 4x-2y+5z+7=0
b) 3x-2y-3z+1=0
c) 4x-y+5z+7=0
d) 4x-2y-z+7=0
View Answer

Answer: a
Explanation: The position vector of the point (3,2,-3) is \(\vec{a}=3\hat{i}+2\hat{j}-3\hat{k}\) and the normal vector \(\vec{N}\) perpendicular to the plane is \(\vec{N}=4\hat{i}-2\hat{j}+5\hat{k}\)
Therefore, the vector equation of the plane is given by \((\vec{r}-\vec{a}).\vec{N}\)=0
Hence, \((\vec{r}-(3\hat{i}+2\hat{j}-3\hat{k})).(4\hat{i}-2\hat{j}+5\hat{k})\)=0
4(x-3)-2(y-2)+5(z+3)=0
4x-2y+5z+7=0.

Sanfoundry Global Education & Learning Series – Mathematics – Class 12.

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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