JEE Challenger
More from Vector Algebra

Vectors Coplanar and Perpendicular to Given Vectors

The vector(s) which is/are coplanar with vectors i^+j^+2k^\hat{i}+\hat{j}+2\hat{k} and i^+2j^+k^\hat{i}+2\hat{j}+\hat{k}, and perpendicular to the vector i^+j^+k^\hat{i}+\hat{j}+\hat{k} is/are

Options

A

j^−k^\hat{j}-\hat{k}

Correct
B

−i^+j^-\hat{i}+\hat{j}

C

i^−j^\hat{i}-\hat{j}

D

−j^+k^-\hat{j}+\hat{k}

Correct

Step-by-Step Solution

To find the vector(s) coplanar with the vectors a⃗=i^+j^+2k^\vec{a} = \hat{i}+\hat{j}+2\hat{k} and b⃗=i^+2j^+k^\vec{b} = \hat{i}+2\hat{j}+\hat{k}, and perpendicular to c⃗=i^+j^+k^\vec{c} = \hat{i}+\hat{j}+\hat{k}, we can express the required vector r⃗\vec{r} as a linear combination of a⃗\vec{a} and b⃗\vec{b}:

r⃗=λa⃗+μb⃗\vec{r} = \lambda \vec{a} + \mu \vec{b}

Substituting the components of a⃗\vec{a} and b⃗\vec{b}:
r⃗=λ(i^+j^+2k^)+μ(i^+2j^+k^)\vec{r} = \lambda(\hat{i}+\hat{j}+2\hat{k}) + \mu(\hat{i}+2\hat{j}+\hat{k})
r⃗=(λ+μ)i^+(λ+2μ)j^+(2λ+μ)k^\vec{r} = (\lambda + \mu)\hat{i} + (\lambda + 2\mu)\hat{j} + (2\lambda + \mu)\hat{k}

Since r⃗\vec{r} is perpendicular to c⃗=i^+j^+k^\vec{c} = \hat{i}+\hat{j}+\hat{k}, their dot product must be zero:
r⃗⋅c⃗=0\vec{r} \cdot \vec{c} = 0
(λ+μ)(1)+(λ+2μ)(1)+(2λ+μ)(1)=0(\lambda + \mu)(1) + (\lambda + 2\mu)(1) + (2\lambda + \mu)(1) = 0
4λ+4μ=0  ⟹  λ+μ=0  ⟹  μ=−λ4\lambda + 4\mu = 0 \implies \lambda + \mu = 0 \implies \mu = -\lambda

Substituting μ=−λ\mu = -\lambda back into the expression for r⃗\vec{r}:
r⃗=(λ−λ)i^+(λ−2λ)j^+(2λ−λ)k^\vec{r} = (\lambda - \lambda)\hat{i} + (\lambda - 2\lambda)\hat{j} + (2\lambda - \lambda)\hat{k}
r⃗=−λj^+λk^=λ(−j^+k^)\vec{r} = -\lambda \hat{j} + \lambda \hat{k} = \lambda(-\hat{j} + \hat{k})

  • For λ=−1\lambda = -1, we get:
    r⃗=j^−k^\vec{r} = \hat{j} - \hat{k}
    which corresponds to option (A).

  • For λ=1\lambda = 1, we get:
    r⃗=−j^+k^\vec{r} = -\hat{j} + \hat{k}
    which corresponds to option (D).

Therefore, the correct options are A and D.