Question

Let $ \vec{a} = \hat{i} + 2 \hat{j} + \hat{k} $ and $ \vec{b} = 2 \hat{i} + \hat{j} - \hat{k} $. Let $ \hat{c} $ be a unit vector in the plane of the vectors $ \vec{a} $ and $ \vec{b} $ and perpendicular to $ \vec{a} $. Then such a vector $ \hat{c} $ is:

• A. \( \frac{1}{\sqrt{3}} (\hat{i} - \hat{j} + \hat{k}) \)
• B. \( \frac{1}{\sqrt{2}} (-\hat{i} + \hat{k}) \)
• C. \( \frac{1}{\sqrt{5}} (\hat{j} - 2\hat{k}) \)
• D. \( \frac{1}{\sqrt{3}} (-\hat{i} + \hat{j} - \hat{k}) \)

Hint:

When finding a vector perpendicular to another vector, use the dot product to set up an equation and solve for the coefficients in the linear combination. Normalize the result to make it a unit vector.

Answer 1

The Correct Option is B

1. Express $\mathbf{c}$ as a linear combination of $\mathbf{a}$ and $\mathbf{b}$
Given that $\mathbf{c}$ lies in the plane spanned by $\mathbf{a}$ and $\mathbf{b}$, it can be represented as:
$\mathbf{c} = x\mathbf{a} + y\mathbf{b}$, where $x$ and $y$ are scalar coefficients.
Substituting the vector components:
$\mathbf{c} = x(i + 2j + k) + y(2i + j - k)$
Combining the components:
$\mathbf{c} = (x + 2y)i + (2x + y)j + (x - y)k$

2. Apply the perpendicularity condition
Since $\mathbf{c}$ is perpendicular to $\mathbf{a}$, their dot product is zero:
$\mathbf{a} \cdot \mathbf{c} = 0$
Performing the dot product:
$(i + 2j + k) \cdot ((x + 2y)i + (2x + y)j + (x - y)k) = 0$
Expanding the dot product:
$(x + 2y) + 2(2x + y) + (x - y) = 0$
Simplifying the equation:
$x + 2y + 4x + 2y + x - y = 0$
$6x + 3y = 0$
Further simplification yields:
$2x + y = 0$
Solving for $y$ in terms of $x$: $y = -2x$

3. Substitute $y$ into the expression for $\mathbf{c}$
Replace $y$ with $-2x$ in the vector equation for $\mathbf{c}$:
$\mathbf{c} = (x + 2(-2x))i + (2x + (-2x))j + (x - (-2x))k$
Performing the substitutions:
$\mathbf{c} = (x - 4x)i + (2x - 2x)j + (x + 2x)k$
Simplifying the components:
$\mathbf{c} = -3xi + 0j + 3xk$
Factoring out $x$: $\mathbf{c} = x(-3i + 3k)$

4. Utilize the unit vector condition
As $\mathbf{c}$ is a unit vector, its magnitude is 1:
$||\mathbf{c}|| = 1$
Calculating the magnitude:
$\sqrt{ (-3x)^2 + (3x)^2} = 1$
$\sqrt{9x^2 + 9x^2} = 1$
$\sqrt{18x^2} = 1$
Simplifying the radical: $3\sqrt{2} |x| = 1$
Solving for $|x|$: $|x| = \frac{1}{3\sqrt{2}}$
This gives two possible values for $x$: $x = \frac{1}{3\sqrt{2}}$ or $x = -\frac{1}{3\sqrt{2}}$

5. Determine the possible vectors $\mathbf{c}$

• Case 1: If $x = \frac{1}{3\sqrt{2}}$:
  $\mathbf{c} = \frac{1}{3\sqrt{2}}(-3i + 3k) = \frac{1}{\sqrt{2}}(-i + k)$
• Case 2: If $x = -\frac{1}{3\sqrt{2}}$:
  $\mathbf{c} = -\frac{1}{3\sqrt{2}}(-3i + 3k) = \frac{1}{\sqrt{2}}(i - k) = -\frac{1}{\sqrt{2}}(-i + k)$

6. Match with the provided options
One of the possible vectors for $\mathbf{c}$ is $\mathbf{c} = \frac{1}{\sqrt{2}}(-i + k)$.
Answer: The correct answer corresponds to option 2.