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Install NowCBSE Class 12 Mathematics Vector Algebra Extra Questions. myCBSEguide has just released Chapter Wise Question Answers for class 12 Maths. There chapter wise Practice Questions with complete solutions are available for download in myCBSEguide website and mobile app. These Questions with solution are prepared by our team of expert teachers who are teaching grade in CBSE schools for years. There are around 4-5 set of solved Chapter 10 Vector Algebra Mathematics Extra Questions from each and every chapter. The students will not miss any concept in these Chapter wise question that are specially designed to tackle Board Exam. We have taken care of every single concept given in CBSE Class 12 Mathematics syllabus and questions are framed as per the latest marking scheme and blue print issued by CBSE for class 12.
Class 12 Chapter 10 Maths Extra Questions
Vector Algebra Extra Questions Class 12 Maths
Chapter 10 Vector Algebra
Find the angle between two vectors →a and →b with magnitudes √3 and 2, respectively, having →a.→b=√6.
- π5
- π3
- π2
- π4
Find the angle between two vectors ˆi–2ˆj+3ˆkand 3ˆi–2ˆj+ˆk.
- cos–1(47)
- cos–1(67)
- cos–1(59)
- cos–1(57)
Vector has
- direction
- None of these
- magnitude
- magnitude as well as direction
Find the sum of the vectors→a=ˆi–2ˆj+ˆk,→b=–2ˆi+4ˆj+5ˆk and →c=ˆi–6ˆj–7ˆk.
- –ˆi+4ˆj–ˆk
- –4ˆj–ˆk
- –ˆi–4ˆj–ˆk
- ˆi–4ˆj–ˆk
Find the direction cosines of the vector ˆi+2ˆj+3ˆk.
- 1√14,2√14,3√14
- 1√14,2√14,–3√14
- 1√14,–2√14,3√14
- –1√14,2√14,3√14
- The values of k which |k→a|<|→a| and k→a+12→a is parallel to →a holds true are ________.
- If →r.→a=0, →r.→b=0, and →r.→c=0 for some non-zero vector →r, then the value of →a(→b×→c) is ________.
- The angle between two vectors →a and →b with magnitudes √3 and 4, respectively, →a.→b = 2√3 is ________.
Find →a×→b if →a=2ˆi+ˆj+3ˆk,→b=3ˆi+5ˆj–2ˆk.
Find the projection of →a on →b, if →a⋅→b=8 and →b=2ˆi+6ˆj+3ˆk.
→a Is unit vector and (→x–→a)(→x+→a)=8, Then find |→x|.
Find the position vector of the mid-point of the vector joining the points P (2, 3, 4) and Q(4,1, – 2)
Find sine of the angle between the vectors. →a=2ˆi–ˆj+3ˆk,→b=ˆi+3ˆj+2ˆk.
Find the projection of the vector ˆi+3ˆj+7ˆk on the vector 7ˆi–ˆj+8ˆk
Let →a=ˆi+ˆj+ˆk,→b=4ˆi–2ˆj+3ˆk and →c=ˆi–2ˆj+ˆk.Find a vector of magnitude 6 units, which is parallel to the vector 2→a–→b+3→c.
Let →a=ˆi+4ˆj+2ˆk,→b=3ˆi–2ˆj+7ˆk and →c=2ˆi–ˆj+4ˆk .Find a vector →d which is perpendicular to both →a and →b and →c.→d=15.
A girl walks 4 km towards west, then she walks 3 km in a direction 300 east of north and stops. Determine the girl’s displacement from her initial point of departure.
Find a vector →d which is ⊥ to both →a and →b and →c. →d=15 Let →a=ˆi+4ˆj+2ˆk,→b=3ˆi–2ˆj+7ˆk and →c=2ˆi–ˆj+4ˆk.
Chapter 10 Vector Algebra
Solution
- π4, Explanation: |→a|=√3,|→b|=2,→a.→b=√6
⇒→a.→b=|→a|.|→b|cosθ⇒√6
=2√3cosθ
⇒cosθ=1√2⇒θ=π4
- cos–1(57), Explanation: →a=ˆi–2ˆj+3ˆk,→b=3ˆi–2ˆj+ˆk⇒|→a|=√14,|→b|=√14,→a.→b=10
⇒→a.→b|→a||→b|=cosθ⇒1014=cosθ
⇒cosθ=57⇒θ=cos–157
- magnitude as well as direction, Explanation: A vector has both magnitude as well as direction.
- –4ˆj–ˆk, Explanation: We have: vectors →a=ˆi−2ˆj+ˆk, →b=−2ˆi+4ˆj+5ˆk and
- 1√14,2√14,3√14, Explanation: Let →a=ˆi+2ˆj+3ˆk,Then, ˆa=→a|→a|=ˆi+2ˆj+3ˆk√12+22+32=ˆi+2ˆj+3ˆk√14
Therefore , the D.C.’s of vector a are :
1√14,2√14,3√14.
- π4, Explanation: |→a|=√3,|→b|=2,→a.→b=√6
- k ∈ ]-1, 1 [k ≠ −12
- 0
- π3
- →a×→b=|ˆiˆjˆk21335–2|
=ˆi(–2–15)–ˆj(–4–9)+ˆk(10–3)
=–17ˆi+13ˆj+7ˆk - We are given that, →a⋅→b=8 and →b=2ˆi+6ˆj+3ˆk
∴ The projection of →a on →b is given as = →a⋅→b|→b|
=8√22+62+32
=8√4+36+9
=8√49=87 - |→a|=1
(→x–→a).(→x+→a)=8
|→x|2−|→a|2=8
|→x|2−1=8
|→x|2=9
|→x|=3 - Given: Point P (2, 3, 4) and Q(4,1, – 2)
∴ Position vector of point P is →a=2ˆi+3ˆj+4ˆk
And Position vector of point Q is →b=4ˆi+ˆj–2ˆk
And Position vector of mid-point R of PQ is →a+→b2=2ˆi+3ˆj+4ˆk+4ˆi+ˆj–2ˆk2
=6ˆi+4ˆj+2ˆk2=3ˆi+2ˆj+ˆk - →a×→b=|ˆiˆjˆk2–13132|
=–11ˆi–ˆj+7ˆk
|→a×→b|=√(–11)2+(–1)2+(7)2
=√171=3√19
sinθ=|→a×→b||→a||→b|=3√19√14.√14=314√19 - Let →a=ˆi+3ˆj+7ˆk and →b=7ˆi–ˆj+8ˆk
Projection of vector →a on →b=→a.→b|→b|
=(1)(7)+(3)(–1)+7(8)√(7)2+(–1)2+(8)2
=7–3+56√49+61+64=60√114 - According to the question ,
→a=ˆi+ˆj+ˆk,
→b=4ˆi–2ˆj+3ˆk and
→c=ˆi–2ˆj+ˆk
Now ,2→a–→b+→3→c
=2(ˆi+ˆj+ˆk)–(4ˆi–2ˆj+3ˆk)+3(ˆi–2ˆj+ˆk)
=2ˆi+2ˆj+2ˆk–4ˆi+2ˆj–3ˆk+3ˆi–6ˆj+3ˆk
=ˆi–2ˆj+2ˆk
⇒2→a–→b+3→c=ˆi–2ˆj+2ˆk
Now, a unit vector in the direction of vector is 2→a–→b+3→c=2→a–→b+3→c|2→a–→b+3→c|
=ˆi–2ˆj+2ˆk√(1)2+(–2)2+(2)2
=ˆi–2ˆj+2ˆk√9
=ˆi–2ˆj+2ˆk3
=13ˆi–23ˆj+23ˆk
Vector of magnitude 6 units parallel to the vector is ,
=6(13ˆi–23ˆj+23ˆk)
=2ˆi–4ˆj+4ˆk - Given: Vectors →a=ˆi+4ˆj+2ˆk and →b=3ˆi–2ˆj+7ˆk
We know that the cross-product of two vectors, →a×→b is a vector perpendicular to both →a and →b
Hence, vector →d which is also perpendicular to both →a and →b is →d=λ(→a×→b) where λ=1 or some other scalar.
Therefore, →d=λ|→i→j→k1423–27|
=λ[ˆi(28+4)–ˆj(7–6)+ˆk(–2–12)]
⇒→d=32λˆi–λˆj–14λˆk…(i)
Now given →c=2ˆi–ˆj+4ˆk and →c.→d=15
→c.→d=15
=2(32λ)+(–1)(–λ)+4(–14λ)=15
⇒64λ+λ–56λ=15
⇒9λ=15
⇒λ=159
⇒λ=53
Putting λ=53 in eq. (i), we get
→d=53[32ˆi–ˆj–14ˆk]
⇒→d=13[160ˆi–5ˆj–70ˆk] - Let the initial point of departure is origin (0, 0) and the girl walks a distance OA = 4 km towards west.
Through the point A, draw a line AQ parallel to a line OP, which is 300 East of North, i.e., in East-North quadrant making an angle of 300 with North.
Again, let the girl walks a distance AB = 3 km along this direction →OQ
∴→OA=4(–→i)=–4ˆi …(i) [∵ Vector →OA is along OX’]
Now, draw BM perpendicular to x – axis.
In ΔAMB by Triangle Law of Addition of vectors,
→AB=→AM+→MB=(AM)ˆi+(MB)ˆi
Dividing and multiplying by AB in R.H.S.,
→AB=ABAMABˆi+ABMBABˆj =3cos60oˆi+3sin60oˆj
⇒AB=312ˆi+3√32ˆi=32ˆi+3√32j …(ii)
∴ Girl’s displacement from her initial point O of departure to final point B,
→OB=→OA+→AB =–4ˆi+(32ˆi+3√22ˆj) =(–4+32)ˆi+3√32ˆj
⇒→OB=–52ˆi+3√32ˆj - →a=ˆi+4ˆj+2ˆk,→b=3ˆi–2ˆj+7ˆk and →c=2ˆi–ˆj+4ˆk
Let →d=xˆi+yˆj+zˆk
ATQ, →d.→a=0,→d.→b=0 and →c.→d=15, then,
x + 4y + 2z = 0 …(1)
3x – 2y + 7z = 0 …(2)
2x – y + 4z = 15 …(3)
On solving equation (1) and (2)
x28+4=y6–7=z–2–12=k
x = 32k, y = -k, z = -14k
Put x, y, z in equation (3)
2(32k) – (-k) + 4(-14k) = 15
64k + k – 56k = 15
9k = 15
k=159
k=53
x=32×53=1603
y=–53
z=–14×53=–703
→d=1603ˆi–53ˆj–703ˆk
Chapter Wise Important Questions Class 12 Maths Part I and Part II
- Relations and Functions
- Inverse Trigonometric Functions
- Matrices
- Determinants
- Continuity and Differentiability
- Application of Derivatives
- Integrals
- Application of Integrals
- Differential Equations
- Vector Algebra
- Three Dimensional Geometry
- Linear Programming
- Probability
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