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Class 10 Mathematics Chapter 6 Test 1 | Vectors in Plane | Punjab Board New Syllabus 2027
Class 10 Math Chapter 6 Test 1 Vectors in Plane
Class 10 Mathematics Chapter 6 Test 1 – Vectors in Plane is prepared according to the Punjab Board New Syllabus 2027 and the latest board-style paper pattern. This Class 10 Math Chapter 6 Test 1 covers important concepts and questions from Vectors in Plane to help students prepare for Punjab Board examinations and chapter-wise mathematics tests.
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Unit 06:
Total Time: 35 min
CHAPTER WISE CLASS TEST
(Test # 1)
Mathematics
Total Marks: 25
A B C D
1
A
B
C
D
2
A
B
C
D
3
A
B
C
D
A B C D
4
A
B
C
D
5
A
B
C
D
6
A
B
C
D
A B C D
7
A
B
C
D
8
A
B
C
D
NOTE: Four possible answers A, B, C and D to each question are given.
The choice which you think is correct, fill that circle in front of that
question with Marker or Pen ink. Cutting or filling two or more circles
will result in zero mark in that question.
Answer Key: 1-C 2-A 3-B 4-A 5-C 6-A 7-C 8-B
Q.1:
(8)
1. What is the value of
|3i + 4j|?
Solution:
For a vector v = xi + yj, the magnitude is |v| = √(x² + y²).
Here x = 3 and y = 4:
|3i + 4j| = √(3² + 4²) = √(9 + 16) = √25
= 5 ⟹ Answer: (C)
2. If
OA = a,
OB = b,
then AB:
Solution:
In triangle OAB, going from O to A and then from A to B is the same as going from O to B (triangle law):
OA + AB = OB
AB = OB − OA = b − a
(Rule: vector AB = position vector of the end point B − position vector of the start point A.)
Answer: (A) b − a
3. Sum of two vectors is:
Solution:
When two vectors are added (e.g. (2i + j) + (i + 3j) = 3i + 4j), the result has both magnitude and direction.
A quantity with magnitude and direction is a vector, so the sum (resultant) of two vectors is again a vector.
Answer: (B) a vector
4. An example of scalar quantity is:
Solution:
A scalar quantity has only magnitude (size); a vector quantity has magnitude and direction.
Force, acceleration and velocity all need a direction, so they are vectors.
Mass (e.g. 5 kg) has only magnitude, so it is a scalar.
Answer: (A) Mass
5. A vector whose initial point is origin O (0,0) is:
Solution:
Null vector: magnitude zero. Unit vector: magnitude 1.
A vector that starts at the origin O(0, 0) and ends at a point P is called the position vector of P, written OP.
Answer: (C) Position vector
6. The position vector of point P(-1,4) with respect to O (0,0) is:
Solution:
The position vector of a point P(x, y) is OP = xi + yj.
Here x = −1 and y = 4:
OP = (−1)i + 4j
= −i + 4j ⟹ Answer: (A)
7. If two sides of triangle represent vectors
u and
v in order,
the third side represents:
Solution:
Triangle law of vector addition: if two sides of a triangle represent two vectors u and v taken in order (the head of u is the tail of v), then the third side, drawn from the tail of u to the head of v, represents their sum u + v.
(Dividing vectors, u / v, is not defined; |u + v| is only a length (scalar), not a vector.)
Answer: (C) u + v
8. If
a = −i + j and
b = 2i + 2j,
then a + b =
__________.
Solution:
Add the i-components together and the j-components together:
a + b = (−i + j) + (2i + 2j)
= (−1 + 2)i + (1 + 2)j
= i + 3j ⟹ Answer: (B)
(PART – I)
Q.2
Write short answers to following questions.
(6×2=12)
(i)
Check that the vectors
u = 8i + 4j
and
v = 2i + j
are parallel or not.
Solution:
Two vectors are parallel if one is a scalar multiple of the other, i.e. u = kv.
Compare the components:
i-components: 8 ÷ 2 = 4 j-components: 4 ÷ 1 = 4
Both ratios are equal (k = 4), so
u = 8i + 4j = 4(2i + j) = 4v
Since u = 4v, the vectors u and v are parallel.
(ii)
Find unit vector opposite in direction of
u = 3i − 2j.
Solution:
Step 1: Find the magnitude: |u| = √(3² + (−2)²) = √(9 + 4) = √13
Step 2: The unit vector in the same direction is
u|u| =
3i − 2j√13
Step 3: The unit vector in the opposite direction is the negative of this:
−3i − 2j√13
= −3i + 2j√13
= −3√13 i +
2√13 j
Check: its magnitude = √(9/13 + 4/13) = √(13/13) = 1 ✓
(iii)
If
a = 4i − j
and
b = −3i + 5j,
then find
3b + 4a.
Solution:
Step 1: 3b = 3(−3i + 5j) = −9i + 15j
Step 2: 4a = 4(4i − j) = 16i − 4j
Step 3: 3b + 4a = (−9i + 15j) + (16i − 4j)
= (−9 + 16)i + (15 − 4)j
3b + 4a = 7i + 11j
(iv)
If
u = i + 5j
and
v = 6i − j,
then find
v − 2u.
Solution:
Step 1: 2u = 2(i + 5j) = 2i + 10j
Step 2: v − 2u = (6i − j) − (2i + 10j)
= (6 − 2)i + (−1 − 10)j
v − 2u = 4i − 11j
(v)
Define a translation vector.
Answer:
A translation vector is a vector that describes a translation, i.e. it tells how far and in which direction every point of a figure is moved (shifted). The figure keeps its size and shape; it is only moved to a new position.
Example: translating the point P(2, 3) by the vector 4i − j (move 4 units right, 1 unit down) gives the image point P′(2 + 4, 3 − 1) = P′(6, 2).
(vi)
If
AB = 4i − 5j
and
DC = 4i − yj
are two Parallel and opposite sides a parallelogram,
then what is the value of y?
Solution:
In parallelogram ABCD, the opposite sides AB and DC are parallel and equal in length, and they point the same way when taken in the order A→B and D→C. So
AB = DC
4i − 5j = 4i − yj
Compare the j-components: −5 = −y
y = 5
(PART – II)
Note: Attempt the following question.
(5)
Q.3:
Show that
a = 5i − 2i
and
b = −10i + 4j
are parallel.
Solution:
⚠ As printed, a = 5i − 2i, which would be just 3i and could not be parallel to b. It looks like a typing error for a = 5i − 2j, so it is solved that way. Please check the paper.
Two vectors are parallel if one is a scalar multiple of the other, i.e. b = ka.
Step 1: Write the given vectors: a = 5i − 2j and b = −10i + 4j
Step 2: Compare the components of b with a:
i-components: −10 ÷ 5 = −2
j-components: 4 ÷ (−2) = −2
Step 3: Both ratios are the same (k = −2), so
b = −10i + 4j = −2(5i − 2j) = −2a
Since b = −2a, the vectors a and b are parallel (they point in opposite directions, and b is twice as long as a). ∎
