Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Vector Algebra question

2024 · 27 Jan · Shift 1 · Q48
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Mathematics
  4. /Vector Algebra
  5. /2024 · 27 Jan · Shift 1 · Q48

Vector Algebra question

2024 · 27 Jan · Shift 1 · Q48

JEE MainMathematicsVector AlgebraMCQ+4 / −1
Let a→=i^+2j^+k^,b→=3(i^−j^+k^)\overrightarrow{\mathrm{a}}=\hat{i}+2 \hat{j}+\hat{k}, \overrightarrow{\mathrm{b}}=3(\hat{i}-\hat{j}+\hat{k})a=i^+2j^​+k^,b=3(i^−j^​+k^). Let c→\overrightarrow{\mathrm{c}}c be the vector such that a→×c→=b→\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{b}}a×c=b and a⃗⋅c⃗=3\vec{a} \cdot \vec{c}=3a⋅c=3. Then a⃗⋅((c⃗×b⃗)−b⃗−c⃗)\vec{a} \cdot((\vec{c} \times \vec{b})-\vec{b}-\vec{c})a⋅((c×b)−b−c) is equal to :
  1. A
    32
  2. B
    36
  3. C
    24
  4. D
    20
View written solutionFree

Correct answer: C

  1. Given vectors

a⃗=i^+2j^+k^=(1,2,1),b⃗=3(i^−j^+k^)=(3,−3,3)\vec a=\hat i+2\hat j+\hat k=(1,2,1), \qquad \vec b=3(\hat i-\hat j+\hat k)=(3,-3,3)a=i^+2j^​+k^=(1,2,1),b=3(i^−j^​+k^)=(3,−3,3)

We are also given

a⃗×c⃗=b⃗,a⃗⋅c⃗=3.\vec a\times \vec c=\vec b, \qquad \vec a\cdot \vec c=3.a×c=b,a⋅c=3.

We need to find

a⃗⋅((c⃗×b⃗)−b⃗−c⃗).\vec a\cdot\big((\vec c\times \vec b)-\vec b-\vec c\big).a⋅((c×b)−b−c).


  1. Break the required expression

Using distributivity of dot product,

a⃗⋅((c⃗×b⃗)−b⃗−c⃗)=a⃗⋅(c⃗×b⃗)−a⃗⋅b⃗−a⃗⋅c⃗.\vec a\cdot\big((\vec c\times \vec b)-\vec b-\vec c\big)=\vec a\cdot(\vec c\times \vec b)-\vec a\cdot\vec b-\vec a\cdot\vec c.a⋅((c×b)−b−c)=a⋅(c×b)−a⋅b−a⋅c.

So we compute these three terms one by one.


  1. Compute a⃗⋅b⃗\vec a\cdot\vec ba⋅b

a⃗⋅b⃗=(1,2,1)⋅(3,−3,3)=1⋅3+2⋅(−3)+1⋅3=3−6+3=0.\vec a\cdot\vec b=(1,2,1)\cdot(3,-3,3)=1\cdot 3+2\cdot(-3)+1\cdot 3=3-6+3=0.a⋅b=(1,2,1)⋅(3,−3,3)=1⋅3+2⋅(−3)+1⋅3=3−6+3=0.

Hence,

a⃗⋅b⃗=0.\vec a\cdot\vec b=0.a⋅b=0.

Also given,

a⃗⋅c⃗=3.\vec a\cdot\vec c=3.a⋅c=3.


  1. Compute a⃗⋅(c⃗×b⃗)\vec a\cdot(\vec c\times\vec b)a⋅(c×b)

Use scalar triple product identity:

a⃗⋅(c⃗×b⃗)=c⃗⋅(b⃗×a⃗).\vec a\cdot(\vec c\times\vec b)=\vec c\cdot(\vec b\times\vec a).a⋅(c×b)=c⋅(b×a).

Since b⃗=a⃗×c⃗\vec b=\vec a\times\vec cb=a×c, we can write

a⃗⋅(c⃗×b⃗)=a⃗⋅(c⃗×(a⃗×c⃗)).\vec a\cdot(\vec c\times\vec b)=\vec a\cdot\big(\vec c\times(\vec a\times\vec c)\big).a⋅(c×b)=a⋅(c×(a×c)).

Now use the vector triple product identity:

x⃗×(y⃗×z⃗)=y⃗(x⃗⋅z⃗)−z⃗(x⃗⋅y⃗).\vec x\times(\vec y\times\vec z)=\vec y(\vec x\cdot\vec z)-\vec z(\vec x\cdot\vec y).x×(y​×z)=y​(x⋅z)−z(x⋅y​).

With x⃗=c⃗, y⃗=a⃗, z⃗=c⃗\vec x=\vec c,\ \vec y=\vec a,\ \vec z=\vec cx=c, y​=a, z=c,

c⃗×(a⃗×c⃗)=a⃗(c⃗⋅c⃗)−c⃗(c⃗⋅a⃗).\vec c\times(\vec a\times\vec c)=\vec a(\vec c\cdot\vec c)-\vec c(\vec c\cdot\vec a).c×(a×c)=a(c⋅c)−c(c⋅a).

Therefore,

a⃗⋅(c⃗×b⃗)=a⃗⋅[a⃗(c⃗⋅c⃗)−c⃗(c⃗⋅a⃗)].\vec a\cdot(\vec c\times\vec b)=\vec a\cdot\left[\vec a(\vec c\cdot\vec c)-\vec c(\vec c\cdot\vec a)\right].a⋅(c×b)=a⋅[a(c⋅c)−c(c⋅a)].

So,

a⃗⋅(c⃗×b⃗)=(a⃗⋅a⃗)(c⃗⋅c⃗)−(a⃗⋅c⃗)2.\vec a\cdot(\vec c\times\vec b)=(\vec a\cdot\vec a)(\vec c\cdot\vec c)-(\vec a\cdot\vec c)^2.a⋅(c×b)=(a⋅a)(c⋅c)−(a⋅c)2.

Now,

a⃗×c⃗=b⃗  ⟹  ∣a⃗×c⃗∣2=∣b⃗∣2.\vec a\times\vec c=\vec b \implies |\vec a\times\vec c|^2=|\vec b|^2.a×c=b⟹∣a×c∣2=∣b∣2.

But

∣a⃗×c⃗∣2=∣a⃗∣2∣c⃗∣2−(a⃗⋅c⃗)2.|\vec a\times\vec c|^2=|\vec a|^2|\vec c|^2-(\vec a\cdot\vec c)^2.∣a×c∣2=∣a∣2∣c∣2−(a⋅c)2.

Hence,

(a⃗⋅a⃗)(c⃗⋅c⃗)−(a⃗⋅c⃗)2=∣b⃗∣2.(\vec a\cdot\vec a)(\vec c\cdot\vec c)-(\vec a\cdot\vec c)^2=|\vec b|^2.(a⋅a)(c⋅c)−(a⋅c)2=∣b∣2.

Thus,

a⃗⋅(c⃗×b⃗)=∣b⃗∣2.\vec a\cdot(\vec c\times\vec b)=|\vec b|^2.a⋅(c×b)=∣b∣2.

Now calculate ∣b⃗∣2|\vec b|^2∣b∣2:

∣b⃗∣2=32+(−3)2+32=9+9+9=27.|\vec b|^2=3^2+(-3)^2+3^2=9+9+9=27.∣b∣2=32+(−3)2+32=9+9+9=27.

So,

a⃗⋅(c⃗×b⃗)=27.\vec a\cdot(\vec c\times\vec b)=27.a⋅(c×b)=27.


  1. Substitute into the required expression

a⃗⋅((c⃗×b⃗)−b⃗−c⃗)=27−0−3=24.\vec a\cdot\big((\vec c\times\vec b)-\vec b-\vec c\big)=27-0-3=24.a⋅((c×b)−b−c)=27−0−3=24.


  1. Check options

The value is

24\boxed{24}24​

So the correct option is C.

PreviousNext

More from Vector Algebra

  • The least positive integral value of α, for which the angle between the vectors αi^−2j^​+2k^ and αi^+2αj^​−2k^ is acute, is ​.2024 · Numerical
  • Let the position vectors of the vertices A,B and C of a triangle be 2i^+2j^​+k^,i^+2j^​+2k^ and 2i^+j^​+2k^ respectively. Let l1​,l2​ and l3​ be…2024 · MCQ
  • The position vectors of the vertices A,B and C of a triangle are 2i^−3j^​+3k^,2i^+2j^​+3k^ and −i^+j^​+3k^ respectively. Let l denotes the length of…2024 · MCQ
  • Let a,b and c be three non-zero vectors such that b and c are non-collinear. If a+5b is collinear with c,b+6c is collinear with a and a+αb+βc=0…2024 · MCQ
  • Let a unit vector u^=xi^+yj^​+zk^ make angles 2π​,3π​ and 32π​ with the vectors 2​1​i^+2​1​k^,2​1​j^​+2​1​k^…2024 · MCQ
  • Let OA=a,OB=12a+4b and OC=b, where O is the origin. If S is the parallelogram with adjacent sides OA and OC, then areaofSareaofthequadrilateralOABC​…2024 · MCQ
  • Let a=a1​i^+a2​j^​+a3​k^ and b=b1​i^+b2​j^​+b3​k^ be two vectors such that ∣a∣=1,a⋅b=2…2024 · MCQ
  • Let a=i^+αj^​+βk^,α,β∈R. Let a vector b be such that the angle between a and b is 4π​ and ∣b∣2=6. If $\vec{a} \cdot \vec{b}=3…2024 · MCQ