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Vector Algebra question

2022 · 28 Jul · Shift 1 · Q25
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  5. /2022 · 28 Jul · Shift 1 · Q25

Vector Algebra question

2022 · 28 Jul · Shift 1 · Q25

JEE MainMathematicsVector AlgebraMCQ+4 / −1
Let the vectors a⃗=(1+t)i^+(1−t)j^+k^,b⃗=(1−t)i^+(1+t)j^+2k^\vec{a}=(1+t) \hat{i}+(1-t) \hat{j}+\hat{k}, \vec{b}=(1-t) \hat{i}+(1+t) \hat{j}+2 \hat{k}a=(1+t)i^+(1−t)j^​+k^,b=(1−t)i^+(1+t)j^​+2k^ and c⃗=ti^−tj^+k^,t∈R\vec{c}=t \hat{i}-t \hat{j}+\hat{k}, t \in \mathbf{R}c=ti^−tj^​+k^,t∈R be such that for α,β,γ∈R,αa⃗+βb⃗+γc⃗=0→⇒α=β=γ=0\alpha, \beta, \gamma \in \mathbf{R}, \alpha \vec{a}+\beta \vec{b}+\gamma \vec{c}=\overrightarrow{0} \Rightarrow \alpha=\beta=\gamma=0α,β,γ∈R,αa+βb+γc=0⇒α=β=γ=0. Then, the set of all values of ttt is :
  1. A
    a non-empty finite set
  2. B
    equal to N\mathbf{N}N
  3. C
    equal to R−{0}\mathbf{R}-\{0\}R−{0}
  4. D
    equal to R\mathbf{R}R
View written solutionFree

Correct answer: C

We need the vectors a⃗,b⃗,c⃗∈R3\vec a,\vec b,\vec c \in \mathbb R^3a,b,c∈R3 to satisfy

αa⃗+βb⃗+γc⃗=0⃗  ⟹  α=β=γ=0.\alpha \vec a+\beta \vec b+\gamma \vec c=\vec 0 \implies \alpha=\beta=\gamma=0.αa+βb+γc=0⟹α=β=γ=0.

This means that the three vectors are linearly independent.

So, we must find all real ttt for which the determinant of the matrix formed by these vectors is non-zero.

1. Write the vectors in component form

Given

a⃗=(1+t, 1−t, 1),b⃗=(1−t, 1+t, 2),c⃗=(t, −t, 1).\vec a=(1+t,\,1-t,\,1),\qquad \vec b=(1-t,\,1+t,\,2),\qquad \vec c=(t,\,-t,\,1).a=(1+t,1−t,1),b=(1−t,1+t,2),c=(t,−t,1).

Form the matrix with these as columns:

M=(1+t1−tt1−t1+t−t121).M= \begin{pmatrix} 1+t & 1-t & t\\ 1-t & 1+t & -t\\ 1 & 2 & 1 \end{pmatrix}.M=​1+t1−t1​1−t1+t2​t−t1​​.

The vectors are linearly independent iff

det⁡(M)≠0.\det(M)\neq 0.det(M)=0.

2. Compute the determinant

Expand along the first row:

det⁡(M)=(1+t)∣1+t−t21∣−(1−t)∣1−t−t11∣+t∣1−t1+t12∣.\det(M)= (1+t) \begin{vmatrix} 1+t & -t\\ 2 & 1 \end{vmatrix} -(1-t) \begin{vmatrix} 1-t & -t\\ 1 & 1 \end{vmatrix} +t \begin{vmatrix} 1-t & 1+t\\ 1 & 2 \end{vmatrix}.det(M)=(1+t)​1+t2​−t1​​−(1−t)​1−t1​−t1​​+t​1−t1​1+t2​​.

Now calculate each minor:

First minor

∣1+t−t21∣=(1+t)(1)−(−t)(2)=1+t+2t=1+3t.\begin{vmatrix} 1+t & -t\\ 2 & 1 \end{vmatrix} =(1+t)(1)-(-t)(2)=1+t+2t=1+3t.​1+t2​−t1​​=(1+t)(1)−(−t)(2)=1+t+2t=1+3t.

Second minor

∣1−t−t11∣=(1−t)(1)−(−t)(1)=1−t+t=1.\begin{vmatrix} 1-t & -t\\ 1 & 1 \end{vmatrix} =(1-t)(1)-(-t)(1)=1-t+t=1.​1−t1​−t1​​=(1−t)(1)−(−t)(1)=1−t+t=1.

Third minor

∣1−t1+t12∣=2(1−t)−1(1+t)=2−2t−1−t=1−3t.\begin{vmatrix} 1-t & 1+t\\ 1 & 2 \end{vmatrix} =2(1-t)-1(1+t)=2-2t-1-t=1-3t.​1−t1​1+t2​​=2(1−t)−1(1+t)=2−2t−1−t=1−3t.

So,

det⁡(M)=(1+t)(1+3t)−(1−t)(1)+t(1−3t).\det(M)=(1+t)(1+3t)-(1-t)(1)+t(1-3t).det(M)=(1+t)(1+3t)−(1−t)(1)+t(1−3t).

Expand:

(1+t)(1+3t)=1+4t+3t2,(1+t)(1+3t)=1+4t+3t^2,(1+t)(1+3t)=1+4t+3t2, −(1−t)=−1+t,-(1-t)=-1+t,−(1−t)=−1+t, t(1−3t)=t−3t2.t(1-3t)=t-3t^2.t(1−3t)=t−3t2.

Adding,

det⁡(M)=(1+4t+3t2)+(−1+t)+(t−3t2)=6t.\det(M)=\left(1+4t+3t^2\right)+(-1+t)+(t-3t^2)=6t.det(M)=(1+4t+3t2)+(−1+t)+(t−3t2)=6t.

Thus,

det⁡(M)=6t.\det(M)=6t.det(M)=6t.

3. Condition for linear independence

For linear independence,

6t≠0⇒t≠0.6t\neq 0 \quad \Rightarrow \quad t\neq 0.6t=0⇒t=0.

Hence the set of all values of ttt is

R−{0}.\mathbb R-\{0\}.R−{0}.

4. Match with the options

  • A: a non-empty finite set   →\;\to→ false
  • B: equal to N\mathbf NN   →\;\to→ false
  • C: equal to R−{0}\mathbf R-\{0\}R−{0}   →\;\to→ true
  • D: equal to R\mathbf RR   →\;\to→ false

Final Answer

The correct option is:

C.\boxed{\text{C}}.C​.

This matches the stored correct answer.

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