JEE Challenger
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MathMedAdv 2011 • Paper 1

Vector in Plane of Two Vectors with Given Projection

Let a⃗=i^+j^+k^\vec{a}=\hat{i}+\hat{j}+\hat{k}, b⃗=i^−j^+k^\vec{b}=\hat{i}-\hat{j}+\hat{k} and c⃗=i^−j^−k^\vec{c}=\hat{i}-\hat{j}-\hat{k} be three vectors. A vector v⃗\vec{v} in the plane of a⃗\vec{a} and b⃗\vec{b}, whose projection on c⃗\vec{c} is 13\frac{1}{\sqrt{3}}, is given by

Options

A

i^−3j^+3k^\hat{i}-3\hat{j}+3\hat{k}

B

−3i^−3j^−k^-3\hat{i}-3\hat{j}-\hat{k}

C

3i^−j^+3k^3\hat{i}-\hat{j}+3\hat{k}

Correct
D

i^+3j^−3k^\hat{i}+3\hat{j}-3\hat{k}

Topics & Concepts

Vector AlgebraVector AdditionDot and Cross Product

Step-by-Step Solution

To find the vector v⃗\vec{v}, we use the given conditions:

  1. Coplanarity: Since v⃗\vec{v} lies in the plane of a⃗=i^+j^+k^\vec{a} = \hat{i}+\hat{j}+\hat{k} and b⃗=i^−j^+k^\vec{b} = \hat{i}-\hat{j}+\hat{k}, it can be expressed as a linear combination of a⃗\vec{a} and b⃗\vec{b}:
    v⃗=λa⃗+μb⃗(where λ,μ∈R)\vec{v} = \lambda \vec{a} + \mu \vec{b} \quad (\text{where } \lambda, \mu \in \mathbb{R})

    Substituting the given vectors:
    v⃗=λ(i^+j^+k^)+μ(i^−j^+k^)\vec{v} = \lambda(\hat{i}+\hat{j}+\hat{k}) + \mu(\hat{i}-\hat{j}+\hat{k})
    v⃗=(λ+μ)i^+(λ−μ)j^+(λ+μ)k^\vec{v} = (\lambda + \mu)\hat{i} + (\lambda - \mu)\hat{j} + (\lambda + \mu)\hat{k}

  2. Projection Condition: The projection of v⃗\vec{v} on c⃗=i^−j^−k^\vec{c} = \hat{i}-\hat{j}-\hat{k} is given as 13\frac{1}{\sqrt{3}}.

    The formula for the projection of v⃗\vec{v} on c⃗\vec{c} is:
    Projection=v⃗⋅c⃗∣c⃗∣\text{Projection} = \frac{\vec{v} \cdot \vec{c}}{|\vec{c}|}

    First, calculate the magnitude of c⃗\vec{c}:
    ∣c⃗∣=12+(−1)2+(−1)2=3|\vec{c}| = \sqrt{1^2 + (-1)^2 + (-1)^2} = \sqrt{3}

    Now, compute the dot product v⃗⋅c⃗\vec{v} \cdot \vec{c}:
    v⃗⋅c⃗=(λ+μ)(1)+(λ−μ)(−1)+(λ+μ)(−1)\vec{v} \cdot \vec{c} = (\lambda + \mu)(1) + (\lambda - \mu)(-1) + (\lambda + \mu)(-1)
    v⃗⋅c⃗=(λ+μ)−(λ−μ)−(λ+μ)=μ−λ\vec{v} \cdot \vec{c} = (\lambda + \mu) - (\lambda - \mu) - (\lambda + \mu) = \mu - \lambda

    Equating the projection to 13\frac{1}{\sqrt{3}}:
    μ−λ3=13  ⟹  μ−λ=1\frac{\mu - \lambda}{\sqrt{3}} = \frac{1}{\sqrt{3}} \implies \mu - \lambda = 1

  3. Finding v⃗\vec{v}:
    From μ−λ=1\mu - \lambda = 1, we can choose suitable scalar values to check against the given options.
    Taking λ=1\lambda = 1, we get μ=2\mu = 2.

    Substituting these values into v⃗\vec{v}:
    v⃗=1(i^+j^+k^)+2(i^−j^+k^)=3i^−j^+3k^\vec{v} = 1(\hat{i}+\hat{j}+\hat{k}) + 2(\hat{i}-\hat{j}+\hat{k}) = 3\hat{i} - \hat{j} + 3\hat{k}

Thus, the required vector is 3i^−j^+3k^3\hat{i} - \hat{j} + 3\hat{k}, which corresponds to option (C).

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