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

2025 · 3 Apr · Shift 1 · Q47
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. /2025 · 3 Apr · Shift 1 · Q47

Vector Algebra question

2025 · 3 Apr · Shift 1 · Q47

JEE MainMathematicsVector AlgebraNumerical+4 / −1
Let a⃗=i^+j^+k^,b⃗=3i^+2j^−k^,c⃗=λj^+μk^\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=3 \hat{i}+2 \hat{j}-\hat{k}, \vec{c}=\lambda \hat{j}+\mu \hat{k}a=i^+j^​+k^,b=3i^+2j^​−k^,c=λj^​+μk^ and d^\hat{d}d^ be a unit vector such that a⃗×d^=b⃗×d^\vec{a} \times \hat{d}=\vec{b} \times \hat{d}a×d^=b×d^ and c⃗⋅d^=1\vec{c} \cdot \hat{d}=1c⋅d^=1. If c⃗\vec{c}c is perpendicular to a⃗\vec{a}a, then ∣3λd^+μc⃗∣2|3 \lambda \hat{d}+\mu \vec{c}|^2∣3λd^+μc∣2 is equal to ‾\underline{\hspace{2cm}}​
Numerical answer
View written solutionFree

Correct answer: 5

  1. Given vectors a⃗=(1,1,1),b⃗=(3,2,−1),c⃗=(0,λ,μ).\vec a=(1,1,1),\quad \vec b=(3,2,-1),\quad \vec c=(0,\lambda,\mu).a=(1,1,1),b=(3,2,−1),c=(0,λ,μ). Also, d^\hat dd^ is a unit vector such that a⃗×d^=b⃗×d^\vec a\times \hat d=\vec b\times \hat da×d^=b×d^ and c⃗⋅d^=1.\vec c\cdot \hat d=1.c⋅d^=1. Further, c⃗⊥a⃗\vec c\perp \vec ac⊥a.

  2. Use the cross product condition From a⃗×d^=b⃗×d^,\vec a\times \hat d=\vec b\times \hat d,a×d^=b×d^, we get (a⃗−b⃗)×d^=0⃗. (\vec a-\vec b)\times \hat d=\vec 0.(a−b)×d^=0. Now, a⃗−b⃗=(1−3,1−2,1−(−1))=(−2,−1,2).\vec a-\vec b=(1-3,1-2,1-(-1))=(-2,-1,2).a−b=(1−3,1−2,1−(−1))=(−2,−1,2). Hence d^\hat dd^ is parallel to (−2,−1,2)(-2,-1,2)(−2,−1,2).

    Since d^\hat dd^ is a unit vector, ∣(−2,−1,2)∣=4+1+4=3,|(-2,-1,2)|=\sqrt{4+1+4}=3,∣(−2,−1,2)∣=4+1+4​=3, so d^=±(−23,−13,23).\hat d=\pm \left(-\frac23,-\frac13,\frac23\right).d^=±(−32​,−31​,32​).

  3. Use the perpendicularity condition c⃗⊥a⃗\vec c\perp \vec ac⊥a Since c⃗⋅a⃗=(0,λ,μ)⋅(1,1,1)=λ+μ=0,\vec c\cdot \vec a=(0,\lambda,\mu)\cdot (1,1,1)=\lambda+\mu=0,c⋅a=(0,λ,μ)⋅(1,1,1)=λ+μ=0, we get μ=−λ.\mu=-\lambda.μ=−λ. Therefore, c⃗=(0,λ,−λ).\vec c=(0,\lambda,-\lambda).c=(0,λ,−λ).

  4. Use the condition c⃗⋅d^=1\vec c\cdot \hat d=1c⋅d^=1 Take d^=(−23,−13,23).\hat d=\left(-\frac23,-\frac13,\frac23\right).d^=(−32​,−31​,32​). Then

    =-\frac{\lambda}{3}-\frac{2\lambda}{3}=-\lambda.$$ Since this equals $1$, $$-\lambda=1\implies \lambda=-1,\quad \mu=1.$$ (If we took the opposite unit vector, we would get $\lambda=1,\mu=-1$; final required value remains same.) So one convenient choice is $$\vec c=(0,-1,1),\quad \hat d=\left(-\frac23,-\frac13,\frac23\right).$$

  5. Compute 3λd^+μc⃗3\lambda\hat d+\mu\vec c3λd^+μc Here λ=−1,μ=1\lambda=-1,\mu=1λ=−1,μ=1, so 3λd^+μc⃗=−3d^+c⃗.3\lambda\hat d+\mu\vec c=-3\hat d+\vec c.3λd^+μc=−3d^+c. Now, −3d^=−3(−23,−13,23)=(2,1,−2).-3\hat d=-3\left(-\frac23,-\frac13,\frac23\right)=(2,1,-2).−3d^=−3(−32​,−31​,32​)=(2,1,−2). Hence −3d^+c⃗=(2,1,−2)+(0,−1,1)=(2,0,−1).-3\hat d+\vec c=(2,1,-2)+(0,-1,1)=(2,0,-1).−3d^+c=(2,1,−2)+(0,−1,1)=(2,0,−1).

  6. Find its squared magnitude ∣3λd^+μc⃗∣2=∣(2,0,−1)∣2=22+02+(−1)2=5.|3\lambda\hat d+\mu\vec c|^2=|(2,0,-1)|^2=2^2+0^2+(-1)^2=5.∣3λd^+μc∣2=∣(2,0,−1)∣2=22+02+(−1)2=5.

  7. Final answer 5\boxed{5}5​

  8. Comparison with stored answer Stored correct answer = 555, which matches the derived answer.

PreviousNext

More from Vector Algebra

  • Let a=i^+2j^​+k^,b=3i^−3j^​+3k^,c=2i^−j^​+2k^ and d be a vector such that b×d=c×d and a⋅d=4. Then ∣(a×d)∣2…2025 · Numerical
  • Consider two vectors u=3i^−j^​ and v=2i^+j^​−λk^,λ>0. The angle between them is given by cos−1(27​5​​). Let v=v1​+v2​​…2025 · MCQ
  • Let the three sides of a triangle ABC be given by the vectors 2i^−j^​+k^,i^−3j^​−5k^ and 3i^−4j^​−4k^. Let G be the centroid of the triangle ABC. Then 6(∣AG∣2+∣BG∣2+∣CG∣2)…2025 · Numerical
  • Let the angle θ,0<θ<2π​ between two unit vectors a^ and b^ be sin−1(965​​). If the vector c=3a^+6b^+9(a^×b^), then the value of…2025 · MCQ
  • Let a and b be the vectors of the same magnitude such that ∣a+b∣−∣a−b∣∣a+b∣+∣a−b∣​=2​+1. Then ∣a∣2∣a+b∣2​ is :2025 · MCQ
  • Let a=i^+2j^​+k^ and b=2i^+j^​−k^. Let c^ be a unit vector in the plane of the vectors a and b and be perpendicular to a. Then such a vector c^…2025 · MCQ
  • Let c be the projection vector of b=λi^+4k^,λ>0, on the vector a=i^+2j^​+2k^. If ∣a+c∣=7, then the area of the parallelogram formed by the vectors b…2025 · Numerical
  • Let a and b be two unit vectors such that the angle between them is 3π​. If λa+2b and 3a−λb are perpendicular to each other, then the number of values of λ in [−1,3]…2025 · MCQ