Linear Algebra Questions and Answers

This set of Linear Algebra Multiple Choice Questions & Answers (MCQs) focuses on “Conversion from Cartesian, Cylindrical and Spherical Coordinates”.

1. Convert Cartesian coordinates (2, 6, 9) to Cylindrical and Spherical Coordinates.
a) (6.32, 71.565., 6.32) and (11, 71.565., 35.097)
b) (6.32, 71.565., 9) and (6.32, 71.565., 35.097)
c) (6.32, 71.565., 6.32) and (6.32, 35.097., 71.565)
d) (6.32, 71.565., 9) and (11, 35.097., 71.565)
View Answer

Answer: d
Explanation: The Cylindrical coordinates is of the form ( ρ, φ, z) where \(ρ=\sqrt{x^2+y^2} \) and \( φ=tan^{-1}⁡(\frac{y}{x}) \) and z=z where (x, y, z) is the Cartesian coordinates. The Spherical coordinates is of the form (r, θ, φ) where \(r=\sqrt{(a^2+b^2 +c^2 )}, \, θ = tan^{-1}⁡(\frac{\sqrt{x^2+y^2}}{z}) \) and \( φ=tan^{-1}⁡(\frac{y}{x})\). Now, substituting the values for x as 2, y as 6 and z as 9, we get the answer as (6.32, 71.565., 9) and (11, 35.097., 71.565.).

2. Convert the (10, 90, 60) coordinates to Cartesian coordinates which are in Spherical coordinates.
a) (5, 8.66, 10)
b) (5, 8.66, 0)
c) (10, 5, 8.66)
d) (0, 5, 8.66)
View Answer

Answer: b
Explanation: The Spherical coordinates is of the form (r, θ, φ) and Cartesian coordinates is of the form (x, y, z) where x = r sin⁡θ cos⁡ϕ and y = rsin⁡θ sin⁡ϕ and z=rcos⁡θ. Now, substituting the values for r as 10, θ as 90, and φ as 60, substituting the values we get
x = 10 sin90 cos60 = 5
y = 10 sin90 sin60 = 8.66
z = 10 cos90 = 0.

3. Let there be a vector X = yz² a_x + zx² a_y + xy² a_z. Find X at P(3,6,9) in cylindrical coordinates.
a) 100 a_x – 398 a_y + 108 a_z
b) 103 a_x – 401 a_y + 109 a_z
c) 105 a_x – 393 a_y + 105 a_z
d) 95 a_x – 395 a_y + 100 a_z
View Answer

Answer: a
Explanation: The formula for converting a vector from Cartesian coordinates to Cylindrical coordinates is \(\begin{bmatrix}
X\rho\\
Xφ\\
Xx\\
\end{bmatrix} \) \(= \begin{bmatrix}
cos⁡φ & sin⁡φ & 0\\
-sin⁡φ & cos⁡φ & 0\\
0 & 0 & 1\\
\end{bmatrix} \) \(= \begin{bmatrix}
Xx\\
Xy\\
Xz\\
\end{bmatrix} \)
Substituting the column matrix in right hand side by the given the vector, and solving the matrix we get a vector in cylindrical coordinates. Now change the point P which is in Cartesian coordinates to Cylindrical coordinates. Now, we should substitute the point P in X thus finding the value of X at P. Hence we get the value 100 a_x – 398 a_y + 108 a_z.

4. Convert the vector P to Cartesian coordinates where P = r a_r + cos⁡θ a_φ.
a) \(\frac{1}{\sqrt{x^2+y^2+z^2}} [(\frac{x}{\sqrt{x^2+y^2+z^2}}-\frac{xyz}{\sqrt{x^2+y^2 }})az + (\frac{y}{\sqrt{x^2+y^2+z^2}}+\frac{xyz}{\sqrt{x^2+y^2}})ay+ \frac{z}{\sqrt{x^2+y^2+z^2}} ax] \)
b) \(\frac{1}{\sqrt{x^2+y^2+z^2}} [(\frac{x}{\sqrt{x^2+y^2+z^2}}-\frac{yz}{\sqrt{x^2+y^2 }})ax + (\frac{y}{\sqrt{x^2+y^2+z^2}}+\frac{xz}{\sqrt{x^2+y^2}})ay+ \frac{z}{\sqrt{x^2+y^2+z^2}} az] \)
c) \(\frac{1}{\sqrt{x^2+y^2+z^2}} [(\frac{x}{\sqrt{x^2+y^2+z^2}}-\frac{y}{\sqrt{x^2+y^2 }})ax + (\frac{y}{\sqrt{x^2+y^2+z^2}}+\frac{z}{\sqrt{x^2+y^2}})ay+ \frac{z}{\sqrt{x^2+y^2+z^2}} az] \)
d) \(\frac{1}{\sqrt{x^2+y^2+z^2}} [(\frac{x}{\sqrt{x^2+y^2+z^2}}-\frac{y}{\sqrt{x^2+y^2 }})ax + (\frac{y}{\sqrt{x^2+y^2+z^2}}+\frac{x}{\sqrt{x^2+y^2}})ay+ \frac{z}{\sqrt{x^2+y^2+z^2}} az] \)
View Answer

Answer: b
Explanation: The formula to convert any vector from Spherical coordinates to Cartesian coordinates is given by \(\begin{bmatrix}
Px\\
Py\\
Pz\\
\end{bmatrix} \) \(= \begin{bmatrix}
sin\theta cos⁡φ & cos\theta cos⁡φ & -sinφ\\
sin\theta sin⁡φ & cos⁡\theta sinφ & cosφ\\
cos\theta & -sin\theta & 0\\
\end{bmatrix} \) \(= \begin{bmatrix}
r\\
0\\
cos\theta\\
\end{bmatrix} \)
after substituting the values of the vector. Now, solving the matrix we get the answer
\(\frac{1}{\sqrt{x^2+y^2+z^2}} [(\frac{x}{\sqrt{x^2+y^2+z^2}}-\frac{yz}{\sqrt{x^2+y^2 }})ax + (\frac{y}{\sqrt{x^2+y^2+z^2}}+\frac{xz}{\sqrt{x^2+y^2}})ay+ \frac{z}{\sqrt{x^2+y^2+z^2}} az] \).

5. Find the distance between two points A(5,60.,0) and B(10,90.,0) where the points are given in Cylindrical coordinates.
a) 4.19 units
b) 5.19 units
c) 6.19 units
d) 7.19 units
View Answer

Answer: c
Explanation: First convert each point which is in cylindrical coordinates to Cartesian coordinates. Now using the formula, distance \( = \sqrt{(x-x’)^2+(y-y’)^2 +(z-z’)^2} \), and substituting the values of x, y, and z in it, we get the required answer as 6.19 units. This sum can also be solved using a direct formula to find distance using two points in Cylindrical coordinates.

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6. What is the value of a_z . a_r?
a) 1
b) cos⁡θ
c) sin⁡θ
d) 0
View Answer

Answer: b
Explanation: We first convert a_r to Cartesian coordinates. Substituting the values in the formula,
\(\begin{bmatrix}
Px\\
Py\\
Pz\\
\end{bmatrix} \) \(= \begin{bmatrix}
sin\theta cos⁡φ & cos\theta cos⁡φ & -sinφ\\
sin\theta sin⁡φ & cos⁡\theta sinφ & cosφ\\
cos\theta & -sin\theta & 0\\
\end{bmatrix} \) \(= \begin{bmatrix}
1\\
0\\
0\\
\end{bmatrix} \), we get cos⁡θ.

7. Convert U = xyz + y + xz into Cylindrical coordinates.
a) zρ³sin⁡φ cos⁡φ + ρsin⁡φ + zρcos⁡φ
b) zρ²sin⁡φ cos⁡φ + ρsin⁡φ + zρcos⁡φ
c) zρ³sin⁡φ cos⁡φ + ρ2sin⁡φ + zρcos⁡φ
d) zρ²sin⁡φ cos⁡φ + ρ2sin⁡φ + zρcos⁡φ
View Answer

Answer: b
Explanation: x=ρcos⁡φ, y=ρsin⁡φ and z=z. Using this formula we should substitute the values for x and y and change it to Cylindrical coordinates.
U = xyz + y + xz
U = ρcos⁡φ ρsin⁡φ z+ ρsin⁡φ + ρcos⁡φz
U = zρ²sin⁡φ cos⁡φ + ρsin⁡φ + zρcos⁡φ.

8. Find the distance between A(10, 30,60) and B(8, 60, 90).
a) 4
b) 5
c) 6
d) 7
View Answer

Answer: c
Explanation: We first have to convert the given points to Cartesian coordinates from Spherical coordinates. Then we have to apply the formula for distance between two points in Cartesian coordinates and find the distance.

9. What is the value of a_r.a_x?
a) sin⁡θ cos⁡φ
b) sin⁡θ sin⁡φ
c) cos⁡θ cos⁡θ
d) cos⁡φ sin⁡θ
View Answer

Answer: a
Explanation: Using the formula to convert from Spherical coordinates to Cartesian coordinates and substituting the value of the vector here in Spherical coordinates,
\(\begin{bmatrix}
Px\\
Py\\
Pz\\
\end{bmatrix} \) \(= \begin{bmatrix}
sin\theta cos⁡φ & cos\theta cos⁡φ & -sinφ\\
sin\theta sin⁡φ & cos⁡\theta sinφ & cosφ\\
cos\theta & -sin\theta & 0\\
\end{bmatrix} \) \(= \begin{bmatrix}
1\\
0\\
0\\
\end{bmatrix} \), and doing dot product we get, sin⁡θ cos⁡φ.

10. Express V in terms of Spherical coordinates where V = x + y²z + z³x.
a) rsin⁡θ cos⁡φ + r³sin⁡θ² sin⁡φ cos⁡θ + r⁴sin⁡θ cos⁡θ³ cos⁡φ
b) rsin⁡θ cos⁡φ + r²sin⁡θ³ sin⁡φ cos⁡θ + r⁴sin⁡θ cos⁡θ³ cos⁡φ
c) rsin⁡θ cos⁡φ + r³sin⁡θ² sin⁡φ cos⁡θ + r⁴sin⁡θ cos⁡θ² cos⁡φ
d) rsin⁡θ cos⁡φ + r³sin⁡θ³ sin⁡φ cos⁡θ + r⁴sin⁡θ cos⁡θ² cos⁡φ
View Answer

Answer: a
Explanation: Using the formula x = r sin⁡θ cos⁡ϕ and y = rsin⁡θ sin⁡ϕ and z=rcos⁡θ and substituting the values of x, y, z we get the desired results.
V = x + y²z + z³x
V = rsin⁡θ cos⁡φ + r³sin⁡θ² sin⁡φ cos⁡θ + r⁴sin⁡θ cos⁡θ³ cos⁡φ.

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