Chapter 10 Vector Algebra | class 12th | Important Question for Maths

Class 12 Maths Chapter 10 Important Extra Questions Vectors

Vectors Important Extra Questions Very Short Answer Type

Question 1.
Classify the following measures as scalar and vector quantities:
(i) 40°
(ii) 50 watt
(iii) 10gm/cm³
(iv) 20 m/sec towards north
(v) 5 seconds. (N.C.E.R.T.)
Solution:
(i) Angle-scalar
(ii) Power-scalar
(iii) Density-scalar
(iv) Velocity-vector
(v) Time-scalar.

Question 2.
In the figure, which of the vectors are:
(i) Collinear
(ii) Equal
(iii) Co-initial. (N.C.E.R.T.)

Solution:
(i) a⃗ ,c⃗  and a⃗  are collinear vectors.
(ii) a⃗  and c⃗  are equal vectors.
(iii) b⃗ , c⃗  and d⃗  are co-initial vectors.

Question 3.
Find the sum of the vectors:
a⃗  = i^−2j^+k^, a⃗  = −2i^+4j⃗ +5k^ and c⃗  = i^−6j^−7k^. (C.B.S.E. 2012)
Solution:
Sum of the vectors = a^+b^+c^

Question 4.
Find the vector joining the points P (2,3,0) and Q (-1, – 2, – 4) directed from P to Q. (N.C.E.R.T.)
Solution:
Since the vector is directed from P to Q,
∴ P is the initial point and Q is the terminal point.
∴ Reqd. vector = PQ−→−

Question 5.
If a⃗ =xi^+2j^−zk^ and b⃗ =3i^−yj^+k^ are two equal vectors, then write the value of x + y + z. (C.B.S.E. 2013)
Solution:
Here
a⃗ =b⃗ ⇒xi^+2j^−zk^=3i^−yj^+k^
Comparing,A: = 3,2 = -y i.e.y = -2,~z= 1 i.e. z = -1.
Hence, x + y + z = 3 – 2 – 1 = 0.

Question 6.
Find the unit vector in the direction of the sum of the vectors:
a⃗ =2i^−j^+2k^ and b⃗ =−i^+j^+3k^ (N.C.E.R.T.)
Solution:

Question 7.
Find the value of ‘p’ for which the vectors : 3i^+2j^+9k^ and i^−2pj^+3k^ are parallel. (A.I.C.B.S.E. 2014)
Solution:
The given vectors 3i^+2j^+9k^ and i^−2pj^+3k^ are parallel
If 31=2−2p=93 if 3 = 1−p = 3
if p = −13

Question 8.
If a⃗  and b⃗  are perpendicular vectors, |a⃗ +b⃗ | = 13 and |a⃗ | =5, find the value of |b⃗ | (A.I.C.B.S.E. 2014)
Solution:

[∵ a⃗  and b⃗  are perpendicular ⇒ a⃗  b⃗  = 0 ]

⇒ |b⃗ |2 = 169 – 25 = 144.
Hence, |b⃗ | = 12.

Question 9.
Find the magnitude of each of the two vectors a⃗  and b⃗  , having the same magnitude such that the angle between them is 60° and their scalar product is 92 (C.B.S.E. 2018)
Solution:
By the question, |a⃗ |=|b⃗ | …(1)
Now a⃗ ⋅b⃗ =|a⃗ ||b⃗ | cos θ
⇒ 92 = |a⃗ ||a⃗ | cos 60° [Using (1)]
⇒ 92=|a⃗ |2(12)
⇒ |a⃗ |2 = 9.
Hence, |a⃗ |=|b⃗ | = 3.

Question 10.
Find the area of the parallelogram whose diagonals are represented by the vectors: a⃗ =2i^−3j^+4k^ and b⃗ =2i^−j^+2k^ (C.B.S.E. Sample Paper 2018-19)
Solution:

= i^(-6 + 4) – j^(4 – 8) + k^( – 2 + 6)
= – 2i^ + 4j^ + 4k^.
∴ |a⃗ ×b⃗ |=4+16+16−−−−−−−−−√=36−−√= 6.
∴ Area of the parallelogram = 12|a⃗ ×b⃗ |
= 12 (6) = 3 sq. units.

Question 11.
Find the angle between the vectors: a⃗ =i^+j^−k^ and b⃗ =i^−j^+k^ (C.B.S.E. Sample Paper 2018-19)
Solution:
We have:a⃗ =i^+j^−k^ and b⃗ =i^−j^+k^
If ‘θ’ be the angle between a⃗  and b⃗ 

Question 12.
Find a⃗ ⋅(b⃗ ×c⃗ ), if : a⃗ =2i^+j^+3k^,b⃗ =−i^+2j^+k^, and c⃗ =3i^+j^+2k^ (A.I.C.B.S.E. 2014)
Solution:
We have:

= 2 (4 – 1) – (1) (-2 – 3) + 3 (-1 – 6)
= 6 + 5-21 = 11 – 21 = -10.

Vectors Important Extra Questions Short Answer Type

Question 1.
If θ is the angle between two vectors:
i^−2j^+3k^ and 3i^−2j^+k^ ,find sinθ.   (C.B.S.E. 2018)
Solution:

Question 2.
X and Y are two points with position vectors 3a−→+b⃗  and a⃗ −3b⃗  respectively. Write the po-sition vector of a point Z which divides the line segment XY in the ratio 2:1 externally. (C.B.S.E. Outside Delhi 2019)
Solution:
Position vector of
A = 2(a⃗ −3b⃗ )−(3a⃗ +b⃗ )2−1 = −a⃗ −7b⃗ 

Question 3.
Find the unit vector perpendicular to both a→ and b→, where:
a⃗ =4i^−j^+8k^ and b⃗ =−j^+k^
(C.B.S.E. 2019 C)
Solution:

Hence, the unit vector perpendicular to both a⃗  and b⃗ 

Question 4.
If a⃗ =2i^+2j^+k^ , b⃗ =−i^+2j^+k^ and c⃗ =3i^+j^ are such that a⃗ +λb⃗  is perpendicular to c⃗ , then find the value of λ . (C.B.S.E. 2019 C)
Solution:
We have:
a = a⃗ =2i^+2j^+k^ and b⃗ =−i^+2j^+k^
∴ a⃗ +λb⃗ =(2i^+2j^+3k^)+λ(−i^+2j^+k^)
= (2 – λ)i^+(2 + 2λ)j^ + (3+λ)k^.
Now, (a⃗ +λb⃗ ) is perpendicular to c ,
∴ (a⃗ +λb⃗ )⋅c⃗ =0
⇒ ((2 – λ) i^ + (2 + 2λ)j^ + (3 + λk^). (3i^ + j^) = 0
⇒ (2 – λ)(3) + (2 + 2λ) (1) + (3 + λ)(0) = 0
⇒ 6 – 3λ + 2 + 2λ = 0
⇒ -λ, + 8 =0.
Hence, λ,=8.

Question 5.
Let a⃗ =i^+2j^−3k^ and b⃗ =3i^−j^+2k^ be two vectors. Show that the vectors (a⃗ +b⃗ ) and
(a⃗ −b⃗ ) are perpendicular to each other. (C.B.S.E. Outside Delhi 2019)
Solution:
Here, a⃗ +b⃗  = (i^ + 2j^ – 3k^) + (3i^ – j^ + 2k^)
= 4i^ + j^ –k^
and a⃗ −b⃗  = (i^ + 2j^ – 3k^) – (3i^ – j + 2k)
= -2i^ + 3j^ – 5k^.
Now
a⃗ +b⃗  . a⃗ −b⃗  = (4i^ + j^ – k^)- (-2\hat{i} + 3 \hat{j} – 5\hat{k})
= (4) (- 2) + (1) (3)+(-1) (- 5)
= – 8 + 3 + 5 = 0.
Hence \vec{a}+\vec{b} is perpendicular to \vec{a}-\vec{b}.

Question 6.
If the sum of two unit vectors is a unit vector, prove that the magnitude of their difference
is √3 . (C.B.S.E. 2019)
Solution:

By the question,
|\overrightarrow{\mathrm{AB}}|=|\overrightarrow{\mathbf{B C}}|=|\overrightarrow{\mathrm{AC}}|=1
⇒ ΔABC is equilateral, each of its angles be¬ing 60°
⇒ ∠DAB = 2 x 60° = 120° and ∠ADB = 30°.
By Sine Formula,
\frac{\mathrm{DB}}{\sin \angle \mathrm{DAB}}=\frac{\mathrm{AB}}{\sin \angle \mathrm{ADB}}

Question 7.
If \vec{a}+\vec{b}+\vec{c}=\overrightarrow{0} and |\vec{a}| = 3, |\vec{b}| = 5 and |\vec{a}| = 7, then find the value of \overrightarrow{a} \cdot \overrightarrow{b}+\overrightarrow{b} \cdot \overrightarrow{c}+\overrightarrow{c} \cdot \overrightarrow{a}
(C.B.S.E. Sample Paper 2019-20)
Solution:

Question 8.
Find |\vec{a}-\vec{b}|, if two vectors a and b are such that |\vec{a}| = 2,|\vec{b}| = 3 and \vec{a} \cdot \vec{b} = 4. (N.C.E.R.T.)
Solution:

Question 9.
Find the work done by the force \overrightarrow{\mathbf{F}}=2 \hat{i}+\hat{j}+\hat{k} acting on a particle, if the particle is displaced from the point with position vector 2 \hat{i}+\hat{j}+2 \hat{k} to the point with position vector 3 \hat{i}+4 \hat{j}+5 \hat{k}
Solution:
Here F = 2 \hat{i} + \hat{j} + \hat{k}
and d = (3 \hat{i} + 4\hat{j} + 5k) – (2 \hat{i} + 2\hat{j} + 2\hat{k} )
– \hat{i} +2\hat{j} + 3\hat{k} .
∴ Work done = \overrightarrow{\mathrm{F}} \cdot \vec{d}
= (2 \hat{i} + \hat{j} + \hat{k} ). ( \hat{i} + 2\hat{j} + 3\hat{k} )
= (2) (1) + (1) (2) + (1) (3)
= 2 + 2 + 3
= 7 units.

Question 10.
Find ‘λ’ if
(2\hat{i} + 6\hat{j} + 14\hat{k} ) x (\hat{i} – λ\hat{j} + 7\hat{k} ) = 0. (A.I.C.B.S.E. 2010)
Solution:
We have:
(2\hat{i} + 6\hat{j} + 14\hat{k}) x (\hat{i} – λ\hat{j} + 7\hat{k} ) = 0
⇒ \hat{i} (42 + 14λ) – \hat{j}(14-14) + \hat{k} (-2λ – 6) = 0
⇒ 14\hat{i} (λ + 3)-2(λ + 3) ifc = o
= λ+3 = 0.
Hence, λ = -3.

Question 11.
If \vec{a} and \vec{b} are two vectors such that |\vec{a} \cdot \vec{b}|
=|\vec{a} \times \vec{b}|, then what is the angle between \vec{a} and b ? (A.I.C.B.S.E. 2010)
Solution:
We have |\vec{a} \cdot \vec{b}|=|\vec{a} \times \vec{b}|
⇒ |\vec{a}||\vec{b}| \cos \theta=|\vec{a}||\vec{b}| \sin \theta
⇒ cos θ = sin θ ⇒ tan θ = 1 ⇒ θ = 45°.
Hence, the angle between \vec{a} and \vec{a} = 45°.

Question 12.
If \vec{a}=2 \hat{i}+3 \hat{j}+\hat{k}, b = \vec{b}=\hat{i}-2 \hat{j}+\hat{k} and \vec{c}=-3 \hat{i}+\hat{j}+2 \hat{k} ,find [\vec{a} \vec{b} \vec{c}].   (C.B.S.E. 2019)
Solution:
We have : \vec{a} = 2\hat{i} + 3\hat{j} + \hat{k}, \vec{b} = \hat{i} – 2\hat{j} + \hat{k} and \vec{c} = – 3\hat{i} + \hat{j} + 2\hat{k}
∴ [\vec{a} \vec{b} \vec{c}]=\left|\begin{array}{rrr} 2 & 3 & 1 \\ 1 & -2 & 1 \\ -3 & 1 & 2 \end{array}\right|
= 2 (-4-1)-3(2 + 3)+ 1.(1 -6)
= 2(-5)-3(5)+ 1(-5)
= -10-15-5 = -30.

Question 13.
For three non-zero vectors \vec{a}, \vec{b} and \vec{c}, prove that [\vec{a}-\vec{b}, \vec{b}-\vec{c}, \vec{c}-\vec{a}] = 0. (C.B.S.E. 2019)
Solution:

Vectors Important Extra Questions Long Answer Type

Question 1.
Let \vec{a}=4 \hat{i}+5 \hat{j}-\hat{k} and \vec{b}=\hat{i}-4 \hat{j}+5 \hat{k} and \vec{c}=3 \hat{i}+\hat{j}-\hat{k}. Find a vector \vec{a} which is perpendicular to both \vec{c} and \vec{b} and
\vec{d} \cdot \vec{a} =21. (C.B.S.E. 2018)
Solution:
Wehave: \vec{a} = 4\hat{i} + 5\hat{j} – \hat{k}
\vec{b} = \hat{i} – 4\hat{j} + 5\hat{k} and
\vec{c} = 3\hat{i} + \hat{j} – \hat{k}
Let \vec{d} = x\hat{i} + y\hat{j} + z\hat{k}

since \vec{d} is perpendicular to both \vec{c} and \vec{b}
\vec{d} \cdot \vec{c} = 0 and \vec{d} \cdot \vec{b} = 0
⇒ (x\hat{i} + y\hat{j} + z\hat{k}) . (3\hat{i} + \hat{j} – \hat{k}) = 0

and (x\hat{i} + y\hat{j} + z\hat{k}) . (\hat{i} -4\hat{j}+5\hat{j}) = 0
⇒ 3x + y-z = 0 …(1)
and x-4y + 5z = 0 …(2)
Also, \vec{d} \cdot \vec{a} = 21
⇒ (x\hat{i} + y\hat{j} + z\hat{k}). (4\hat{i} + 5\hat{j} – \hat{k}) =21
⇒ 4x + 5y-z = 21 …(3)
Multiplying (1) by 5,
1 5x + 5y – 5z = 0 …(4)
Adding (2) and (4),
16x + y = 0 …(5)
Subtracting (1) from (3),
x + 4y = 21 …(6)
From (5),
y = -16x …(7)
Putting in (6),
x – 64x = 21
-63x: = 21
Putting in (7), y = -16\left(-\frac{1}{3}\right)=\frac{16}{3}
Putting in (1), 3\left(-\frac{1}{3}\right)+\frac{16}{3} – z = 0
z = 13/3
Hence \vec{d}=-\frac{1}{3} \hat{i}+\frac{16}{3} \hat{j}+\frac{13}{3} \hat{k}

Question 2.
If \vec{p}=\hat{i}+\hat{j}+\hat{k} and \vec{q}=\hat{i}-2 \hat{j}+\hat{k} ,find a vector of magnitude 5√3 units perpendicular to the vector \vec{q}. and coplanar with vector \vec{p} and \vec{q}. (C.B.S.E. 2018)
Solution:
Let \vec{r}=a \hat{i}+b \hat{j}+c \hat{k} be the vector.
Since \vec{r} \perp \vec{q}
(1) (a) + (-2) (b) + 1 (c) = 0
⇒ a – 2b + c = 0
Again, \vec{p} , \vec{q} and \vec{r} and coplanar,

⇒ (1) (-2c – b) – (1) (c – a) + (1) (b + 2a) = 0
⇒ -2c-b-c + a + b + 2a = 0
⇒ 3a – 3c = 0
⇒ a – c – 0
Solving (1) and (2),