Computer Graphics Questions & Answers – 3D Translation and Scaling

This set of Computer Graphics Multiple Choice Questions & Answers (MCQs) focuses on “3D Translation and Scaling”.

1. Which of the following operation can be applied on a 3 D object to move it along any axis from its original position?
a) Translation
b) Scaling
c) Rotation
d) Shearing
View Answer

Answer: a
Explanation: The process of translation is applied to move a point along a specified axis from its original position in the coordinate plane. Translation of a point adds a constant vector to that point to move it from its original place.

2. If a point (x, y, z) is to be translated by an amount dx, dy and dz respectively, then what will be the value of the new translated points (x1, y1, z1)?
a) x1 = x
y1 = y
z1 = z
b) x1 = dx
y1 = dy
z1 = dz
c) x1 = x + dx
y1 = y + dy
z1 = z + dz
d) x1 = x – dx
y1 = y – dy
z1 = z – dz
View Answer

Answer: c
Explanation: If point (x, y, z) is to be translated by amount dx, dy and dz to a new location (x1, y1, z1) then new coordinates can be obtained by adding dx to x, dy to y and dz to z as:
x1 = dx + x
y1 = dy + y
z1 = dz + z

3. In the equation x1 = dx + x; which part of the equation is called as the translation factor?
a) x1
b) dx
c) x
d) dx + x
View Answer

Answer: b
Explanation: In the equation x1 = dx + x; dx, is the translation factor, i.e. the amount by which the point will be translated. x is the original point. x1 is the coordinate of the point after translation.
advertisement
advertisement

4. If the original point P = (5, 7, 3) and the translation factor, T = (-2, -1, 3), then what will be the value of the final point P1?
a) P1 = (7, 8, 6)
b) P1 = (3, 6, 0)
c) P1 = (7, 8, 0)
d) P1 = (3, 6, 6)
View Answer

Answer: d
Explanation: The new coordinates can be obtained by adding dx to x, dy to y and dz to z as:
x1 = dx + x ; y1 = dy + y ; z1 = dz + z
So, P1 = (5+(-2), 7+(-1), 3+3)
P1 = (3, 6, 6)

5. How many different types of translation are present in computer graphics?
a) 1
b) 2
c) 3
d) 4
View Answer

Answer: c
Explanation: In computer graphics, there are 3 types of translation techniques namely – the point translation (deals with a single point), the line translation (deals with a line) and the polygon translation (deals with a shape).
Sanfoundry Certification Contest of the Month is Live. 100+ Subjects. Participate Now!

6. Which of the following matrix equation is correct for 3 D translation?
a) \(\begin{matrix}
x1 & 1 \; 0 \; 0 \; Tx & x0\\
[ y1 ] = & [ 0 \; 1 \; 0 \; Ty ] & [ y0 ]\\
z1 & 0 \; 0 \; 1 \; Tz & z0\\
1 & 0 \; 0 \; 0 \; 1 & 1
\end{matrix}
\)

b) \(\begin{matrix}
x1 & 1 \; 0 \; 0 \; 1 & x0\\
[ y1 ] = & [ 0 \; 1 \; 0 \; Tx ] & [ y0 ]\\
z1 & 0 \; 0 \; 1 \; Ty & z0\\
1 & 0 \; 0 \; 0 \; Tz & 1
\end{matrix}
\)

c) \(\begin{matrix}
x1 & 1 \; 0 \; 0 \; Tx & x0\\
[ y1 ] = & [ 0 \; 1 \; 0 \; Ty ] & [ y0 ]\\
z1 & 0 \; 0 \; 1 \; Tz & z0\\
1 & 0 \; 0 \; 0 \; 1 & 1
\end{matrix}
\)

d) \(\begin{matrix}
x1 & 0 \; 0 \; 0 \; Tx & x0\\
[ y1 ] = & [ 0 \; 1 \; 0 \; Ty ] & [ y0 ]\\
z1 & 0 \; 1 \; 0 \; Tz & z0\\
1 & 1 \; 0 \; 0 \; 0 & 1
\end{matrix}
\)
View Answer

Answer: a
Explanation: The correct matrix equation for 3 D translation is –
\(\begin{matrix}
x1 & 1 \; 0 \; 0 \; Tx & x0\\
[ y1 ] = & [ 0 \; 1 \; 0 \; Ty ] & [ y0 ]\\
z1 & 0 \; 0 \; 1 \; Tz & z0\\
1 & 0 \; 0 \; 0 \; 1 & 1
\end{matrix}
\)
First matrix is for the final point, second matrix is for the translation factor and the Third matrix is for the original point.

7. Which of the following operation can be applied on a 3 D object to zoom it in or out about any axis from its original position?
a) Translation
b) Scaling
c) Rotation
d) Shearing
View Answer

Answer: b
Explanation: In computer graphics, scaling is a process of modifying or altering the size of objects. Scaling may be used to increase or reduce the size of object.
advertisement

8. What will be the value of new co-ordinates if the old co-ordinates are (X0 = 2, Y0 = 0, Z0 = 4) and the scaling factor is (Sx = 2, Sy = 1, Sz = 3)?
a) (X1 = 4, Y1 = 1, Z1 = 7)
b) (X1 = 0, Y1 = -1, Z1 = 1)
c) (X1 = 1, Y1 = 0, Z1 = 4/3)
d) (X1 = 4, Y1 = 0, Z1 = 12)
View Answer

Answer: d
Explanation: The new co-ordinates would be – (X1 = 4, Y1 = 0, Z1 = 12)
Applying the scaling equations, we have-
X1 = X0 x Sx = 2 x 2 = 4
Y1 = Y0 x Sy = 0 x 1 = 0
Z1 = Z0 x Sz = 4 x 3 = 12

9. Which of the following matrix equation is correct for 3 D scaling of an object?
a) \(\begin{matrix}
x1 & Sx \; 0 \; 0 \; 0 & x0\\
[ y1 ] = & [ 0 \; Sy \; 0 \; 0 ] & [ y0 ]\\
z1 & 0 \; 0 \; Sz \; 0 & z0\\
1 & 0 \; 0 \; 0 \; 1 & 1
\end{matrix}
\)

b) \(\begin{matrix}
x1 & Sx \; 0 \; 0 \; 1 & x0\\
[ y1 ] = & [ 0 \; Sy \; 1 \; 0 ] & [ y0 ]\\
z1 & 0 \; 1 \; Sz \; 0 & z0\\
1 & 1 \; 0 \; 0 \; 0 & 1
\end{matrix}
\)

c) \(\begin{matrix}
x1 & Sx \; 0 \; 0 \; 0 & x0\\
[ y1 ] = & [ 0 \; Sy \; 0 \; 0 ] & [ y0 ]\\
z1 & 0 \; 0 \; Sz \; 0 & z0\\
0 & 0 \; 0 \; 0 \; 1 & 0
\end{matrix}
\)

d) \(\begin{matrix}
x1 & 0 \; 0 \; 0 \; Sx & x0\\
[ y1 ] = & [ 0 \; 0 \; Sy \; 0 ] & [ y0 ]\\
z1 & 0 \; Sz \; 0 \; 0 & z0\\
1 & 1 \; 0 \; 0 \; 0 & 1
\end{matrix}
\)
View Answer

Answer: a
Explanation: The correct matrix equation for 3 D scaling of an object is –
\(\begin{matrix}
x1 & Sx \; 0 \; 0 \; 0 & x0\\
[ y1 ] = & [ 0 \; Sy \; 0 \; 0 ] & [ y0 ]\\
z1 & 0 \; 0 \; Sz \; 0 & z0\\
1 & 0 \; 0 \; 0 \; 1 & 1
\end{matrix}
\)
First matrix is for the final point, second matrix is for the scaling factor and the third matrix is for the original point.
advertisement

10. If Scaling factor is lesser than 1 then the object size is increased.
a) True
b) False
View Answer

Answer: b
Explanation: The scaling factor of an object is used to determine whether the object is to scaled up or scaled down. If the scaling factor of the object is greater than 1 then the object is increased in scale and if the scaling factor of the object is lesser than 1 then the object is decreased in scale.

Sanfoundry Global Education & Learning Series – Computer Graphics.

To practice all areas of Computer Graphics, here is complete set of 1000+ Multiple Choice Questions and Answers.

If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected]

advertisement
advertisement
Subscribe to our Newsletters (Subject-wise). Participate in the Sanfoundry Certification contest to get free Certificate of Merit. Join our social networks below and stay updated with latest contests, videos, internships and jobs!

Youtube | Telegram | LinkedIn | Instagram | Facebook | Twitter | Pinterest
Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

Subscribe to his free Masterclasses at Youtube & discussions at Telegram SanfoundryClasses.