Question

A vector $\vec{v}$ in the plane of $\vec{a}$ and $\vec{b}$, whose projection on $\vec{c}$ is $\frac{1}{\sqrt{6}}$, is given by

• A. $-\frac{3}{2}\hat{j} + \frac{1}{2}\hat{k}$
• B. $\frac{3}{2}\hat{j} + \frac{1}{2}\hat{k}$
• C. $-\frac{3}{2}\hat{j} - \frac{1}{2}\hat{k}$
• D. $\frac{1}{7}(9\hat{i} + 6\hat{j} + 10\hat{k})$

Answer 1

Answer: (D)

$\frac{1}{7}(9\hat{i} + 6\hat{j} + 10\hat{k})$

Answer 2

The problem asks us to find a vector $\vec{v}$ that satisfies two conditions:

1. $\vec{v}$ lies in the plane of vectors $\vec{a}$ and $\vec{b}$.
2. The projection of $\vec{v}$ on vector $\vec{c}$ is $\frac{1}{\sqrt{6}}$.

Given vectors are: $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ $\vec{b} = 2\hat{i} - \hat{j} + 3\hat{k}$ $\vec{c} = \hat{i} - 2\hat{j} + \hat{k}$

Step 1: Express $\vec{v}$ based on the first condition.

If $\vec{v}$ lies in the plane of $\vec{a}$ and $\vec{b}$ (assuming $\vec{a}$ and $\vec{b}$ are non-collinear, which they are, as $\vec{a} \neq k\vec{b}$), then $\vec{v}$ can be written as a linear combination of $\vec{a}$ and $\vec{b}$: $\vec{v} = x\vec{a} + y\vec{b}$ where $x$ and $y$ are scalar constants.

Substitute the expressions for $\vec{a}$ and $\vec{b}$: $\vec{v} = x(\hat{i} + \hat{j} + \hat{k}) + y(2\hat{i} - \hat{j} + 3\hat{k})$ $\vec{v} = (x + 2y)\hat{i} + (x - y)\hat{j} + (x + 3y)\hat{k}$

Step 2: Use the second condition to find a relationship between $x$ and $y$.

The projection of vector $\vec{v}$ on vector $\vec{c}$ is given by the formula: $Proj_{\vec{c}}\vec{v} = \frac{\vec{v} \cdot \vec{c}}{|\vec{c}|}$

First, calculate the magnitude of $\vec{c}$: $|\vec{c}| = \sqrt{(1)^2 + (-2)^2 + (1)^2} = \sqrt{1 + 4 + 1} = \sqrt{6}$

Next, calculate the dot product $\vec{v} \cdot \vec{c}$: $\vec{v} \cdot \vec{c} = ((x + 2y)\hat{i} + (x - y)\hat{j} + (x + 3y)\hat{k}) \cdot (\hat{i} - 2\hat{j} + \hat{k})$ $\vec{v} \cdot \vec{c} = (x + 2y)(1) + (x - y)(-2) + (x + 3y)(1)$ $\vec{v} \cdot \vec{c} = x + 2y - 2x + 2y + x + 3y$ $\vec{v} \cdot \vec{c} = (x - 2x + x) + (2y + 2y + 3y)$ $\vec{v} \cdot \vec{c} = 0x + 7y = 7y$

Now, substitute these into the projection formula and set it equal to the given value: $\frac{7y}{\sqrt{6}} = \frac{1}{\sqrt{6}}$ Multiplying both sides by $\sqrt{6}$, we get: $7y = 1 \implies y = \frac{1}{7}$

Step 3: Substitute the value of $y$ back into the expression for $\vec{v}$. $\vec{v} = x\vec{a} + \frac{1}{7}\vec{b}$ $\vec{v} = x(\hat{i} + \hat{j} + \hat{k}) + \frac{1}{7}(2\hat{i} - \hat{j} + 3\hat{k})$ $\vec{v} = (x + \frac{2}{7})\hat{i} + (x - \frac{1}{7})\hat{j} + (x + \frac{3}{7})\hat{k}$

Step 4: Check the given options to find the correct value of $x$.

We need to find which of the given options matches this form of $\vec{v}$ for a consistent value of $x$.

Let's test Option 4: $\vec{v} = \frac{1}{7}(9\hat{i} + 6\hat{j} + 10\hat{k}) = \frac{9}{7}\hat{i} + \frac{6}{7}\hat{j} + \frac{10}{7}\hat{k}$ Equating the components with our derived form of $\vec{v}$:

For the $\hat{i}$ component: $x + \frac{2}{7} = \frac{9}{7} \implies x = \frac{9}{7} - \frac{2}{7} = \frac{7}{7} = 1$ For the $\hat{j}$ component: $x - \frac{1}{7} = \frac{6}{7} \implies x = \frac{6}{7} + \frac{1}{7} = \frac{7}{7} = 1$ For the $\hat{k}$ component: $x + \frac{3}{7} = \frac{10}{7} \implies x = \frac{10}{7} - \frac{3}{7} = \frac{7}{7} = 1$

Since all three components yield a consistent value of $x=1$, Option 4 is the correct vector. The vector is $\vec{v} = 1 \cdot \vec{a} + \frac{1}{7} \cdot \vec{b}$. This vector inherently lies in the plane of $\vec{a}$ and $\vec{b}$ by its construction.

Let's verify the vector $\vec{v} = \vec{a} + \frac{1}{7}\vec{b}$: $\vec{v} = (\hat{i} + \hat{j} + \hat{k}) + \frac{1}{7}(2\hat{i} - \hat{j} + 3\hat{k})$ $\vec{v} = \hat{i} + \hat{j} + \hat{k} + \frac{2}{7}\hat{i} - \frac{1}{7}\hat{j} + \frac{3}{7}\hat{k}$ $\vec{v} = (1 + \frac{2}{7})\hat{i} + (1 - \frac{1}{7})\hat{j} + (1 + \frac{3}{7})\hat{k}$ $\vec{v} = \frac{9}{7}\hat{i} + \frac{6}{7}\hat{j} + \frac{10}{7}\hat{k} = \frac{1}{7}(9\hat{i} + 6\hat{j} + 10\hat{k})$

This confirms that Option 4 is the correct vector.