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

Step 1: Use coplanarity.
Since $\vec d$ lies in the plane of $\vec a$ and $\vec b$, write $\vec d=\lambda\vec a+\mu\vec b$ with $\vec a=(3,-1,-1)$, $\vec b=(1,1,-2)$.
Step 2: Write the components.
So $\vec d=(3\lambda+\mu,\,-\lambda+\mu,\,-\lambda-2\mu)$.
Step 3: Apply perpendicular to $\vec c$.
With $\vec c=(2,2,1)$, set $\vec d\cdot\vec c=0$: $2(3\lambda+\mu)+2(-\lambda+\mu)+(-\lambda-2\mu)=0$, which simplifies to $3\lambda+2\mu=0$, so $\mu=-\dfrac{3\lambda}{2}$.
Step 4: Substitute back.
The components become $\dfrac{3\lambda}{2}$, $-\dfrac{5\lambda}{2}$, $2\lambda$, so $\vec d=\dfrac{\lambda}{2}(3,-5,4)$.
Step 5: Use the length $\sqrt2$.
The length of $(3,-5,4)$ is $\sqrt{9+25+16}=\sqrt{50}=5\sqrt2$. So $\left|\dfrac{\lambda}{2}\right|\cdot5\sqrt2=\sqrt2$, giving $|\lambda|=\dfrac25$.
Step 6: Write $\vec d$.
Thus $\vec d=\pm\dfrac15(3\hat i-5\hat j+4\hat k)$. \[ \boxed{\pm\frac15(3\hat i-5\hat j+4\hat k)} \]