Question

Let \( \vec{a} = 3\hat{i} - \hat{j} - \hat{k}, \vec{b} = \hat{i} + \hat{j} - 2\hat{k} \) and \( \vec{c} = 2\hat{i} + 2\hat{j} + \hat{k} \). Let \( \vec{d} \) be a vector such that \( |\vec{d}| = \sqrt{2} \) units. If the vector \( \vec{d} \) is coplanar with \( \vec{a}, \vec{b} \) and perpendicular to \( \vec{c} \), then \( \vec{d} = \)

• A. \( \pm \dfrac{1}{5}(3\hat{i}-5\hat{j}+4\hat{k}) \)
• B. \( \pm \dfrac{1}{5}(-4\hat{i}+5\hat{j}-3\hat{k}) \)
• C. \( \pm \dfrac{1}{5}(3\hat{i}+5\hat{j}-4\hat{k}) \)
• D. \( \pm \dfrac{1}{5}(-3\hat{i}+5\hat{j}+4\hat{k}) \)

Hint:

Whenever a vector is:

• Coplanar with two vectors, express it as their linear combination.

• Perpendicular to another vector, use the dot product equal to zero.

• Given with magnitude, apply the modulus formula at the final step.
This combination of conditions is extremely common in vector algebra problems.

Answer 1

The Correct Option is A

Concept: A vector is said to be coplanar with two vectors if it can be expressed as their linear combination. Also, if a vector is perpendicular to another vector, then their dot product must be zero. The important vector concepts used in this problem are:

• Coplanarity of vectors

• Dot product condition for perpendicular vectors

• Magnitude of a vector

• Linear combination of vectors
We are given: \[ \vec{a}=3\hat{i}-\hat{j}-\hat{k} \] \[ \vec{b}=\hat{i}+\hat{j}-2\hat{k} \] \[ \vec{c}=2\hat{i}+2\hat{j}+\hat{k} \] We need to determine the vector \( \vec{d} \) satisfying all the given conditions.

Step 1: Assume the required vector as a linear combination of \( \vec{a} \) and \( \vec{b} \).
Since \( \vec{d} \) is coplanar with \( \vec{a} \) and \( \vec{b} \), we write: \[ \vec{d}=\lambda \vec{a}+\mu \vec{b} \] Substituting the vectors: \[ \vec{d}=\lambda(3,-1,-1)+\mu(1,1,-2) \] Therefore, \[ \vec{d}=(3\lambda+\mu,\,-\lambda+\mu,\,-\lambda-2\mu) \]

Step 2: Use the perpendicularity condition with vector \( \vec{c} \).
Since \( \vec{d} \perp \vec{c} \), \[ \vec{d}\cdot \vec{c}=0 \] Now, \[ (3\lambda+\mu,\,-\lambda+\mu,\,-\lambda-2\mu)\cdot(2,2,1)=0 \] Expanding the dot product: \[ 2(3\lambda+\mu)+2(-\lambda+\mu)+(-\lambda-2\mu)=0 \] Simplifying carefully: \[ 6\lambda+2\mu-2\lambda+2\mu-\lambda-2\mu=0 \] \[ 3\lambda+2\mu=0 \] Hence, \[ 2\mu=-3\lambda \] \[ \mu=-\frac{3\lambda}{2} \]

Step 3: Substitute the relation into the vector expression.
Substituting \( \mu=-\dfrac{3\lambda}{2} \): First component: \[ 3\lambda+\mu = 3\lambda-\frac{3\lambda}{2} = \frac{3\lambda}{2} \] Second component: \[ -\lambda+\mu = -\lambda-\frac{3\lambda}{2} = -\frac{5\lambda}{2} \] Third component: \[ -\lambda-2\mu = -\lambda+3\lambda = 2\lambda \] Thus, \[ \vec{d} = \left( \frac{3\lambda}{2}, -\frac{5\lambda}{2}, 2\lambda \right) \] Factoring out \( \dfrac{\lambda}{2} \): \[ \vec{d} = \frac{\lambda}{2}(3,-5,4) \] Therefore, \[ \vec{d} = \frac{\lambda}{2}(3\hat{i}-5\hat{j}+4\hat{k}) \]

Step 4: Use the magnitude condition \( |\vec{d}|=\sqrt{2} \).
Magnitude of vector \( (3,-5,4) \): \[ \sqrt{3^2+(-5)^2+4^2} = \sqrt{9+25+16} = \sqrt{50} = 5\sqrt{2} \] Hence, \[ |\vec{d}| = \left|\frac{\lambda}{2}\right|(5\sqrt{2}) \] Given: \[ |\vec{d}|=\sqrt{2} \] Therefore, \[ \left|\frac{\lambda}{2}\right|(5\sqrt{2}) = \sqrt{2} \] Cancelling \( \sqrt{2} \): \[ \frac{5|\lambda|}{2}=1 \] \[ |\lambda|=\frac{2}{5} \] Thus, \[ \lambda=\pm \frac{2}{5} \] Substituting into the vector: \[ \vec{d} = \pm \frac{1}{5}(3\hat{i}-5\hat{j}+4\hat{k}) \] \[ \boxed{ \vec{d} = \pm \frac{1}{5}(3\hat{i}-5\hat{j}+4\hat{k}) } \] Hence, the correct option is: \[ \boxed{(A)} \]