Let $\vec{\alpha}=4 \hat{i}+3 \hat{j}+5 \hat{k}$ and $\vec{\beta}=\hat{i}+2 \hat{j}-4 \hat{k}$ Let $\vec{\beta}_1$ be parallel to $\vec{\alpha}$ and $\vec{\beta}_2$ be perpendicular to $\vec{\alpha}$ If $\vec{\beta}=\vec{\beta}_1+\vec{\beta}_2$, then the value of $5 \vec{\beta}_2 \cdot(\hat{i}+\hat{j}+\hat{k})$ is

Question

Let $ \mathbf{a} $, $ \mathbf{b} $, and $ \mathbf{c} $ be vectors of magnitude 2, 3, and 4 respectively. If: - $ \mathbf{a} $ is perpendicular to $ (\mathbf{b} + \mathbf{c}) $, - $ \mathbf{b} $ is perpendicular to $ (\mathbf{c} + \mathbf{a}) $, - $ \mathbf{c} $ is perpendicular to $ (\mathbf{a} + \mathbf{b}) $, then the magnitude of $ \mathbf{a} + \mathbf{b} + \mathbf{c} $ is equal to:

Answer 1

To find the value of $5 \vec{\beta}_2 \cdot (\hat{i} + \hat{j} + \hat{k})$, we need to resolve the vector $\vec{\beta}$ into components $\vec{\beta}_1$ and $\vec{\beta}_2$, where $\vec{\beta}_1$ is parallel to $\vec{\alpha}$ and $\vec{\beta}_2$ is perpendicular to $\vec{\alpha}$.

Step 1: Determine $\vec{\beta}_1$, the component of $\vec{\beta}$ parallel to $\vec{\alpha}$.

The formula for the projection of $\vec{\beta}$ on $\vec{\alpha}$ is given by:

$\vec{\beta}_1 = \frac{\vec{\alpha} \cdot \vec{\beta}}{\vec{\alpha} \cdot \vec{\alpha}} \vec{\alpha}$

First, calculate $\vec{\alpha} \cdot \vec{\beta}$:

$\vec{\alpha} \cdot \vec{\beta} = (4 \hat{i} + 3 \hat{j} + 5 \hat{k}) \cdot (\hat{i} + 2 \hat{j} - 4 \hat{k})$ = 4 \cdot 1 + 3 \cdot 2 + 5 \cdot (-4) = 4 + 6 - 20 = -10\)

Next, calculate $\vec{\alpha} \cdot \vec{\alpha}$:

$\vec{\alpha} \cdot \vec{\alpha} = (4 \hat{i} + 3 \hat{j} + 5 \hat{k}) \cdot (4 \hat{i} + 3 \hat{j} + 5 \hat{k})$ = 4^2 + 3^2 + 5^2 = 16 + 9 + 25 = 50\)

Calculate $\vec{\beta}_1$:

$\vec{\beta}_1 = \frac{-10}{50} \vec{\alpha} = -\frac{1}{5} (4 \hat{i} + 3 \hat{j} + 5 \hat{k}) = -\frac{4}{5} \hat{i} - \frac{3}{5} \hat{j} - \frac{5}{5} \hat{k}$

Step 2: Determine $\vec{\beta}_2$, the component of $\vec{\beta}$ perpendicular to $\vec{\alpha}$.

Using the relation $\vec{\beta} = \vec{\beta}_1 + \vec{\beta}_2$:

$\vec{\beta}_2 = \vec{\beta} - \vec{\beta}_1$

Substitute the known vectors:

$\vec{\beta}_2 = (\hat{i} + 2 \hat{j} - 4 \hat{k}) - \left(-\frac{4}{5} \hat{i} - \frac{3}{5} \hat{j} - \hat{k}\right)$ = \left(1 + \frac{4}{5}\right) \hat{i} + \left(2 + \frac{3}{5}\right) \hat{j} + \left(-4 + 1\right) \hat{k}\) = \frac{9}{5} \hat{i} + \frac{13}{5} \hat{j} - 3 \hat{k}\)

Step 3: Calculate $5 \vec{\beta}_2 \cdot (\hat{i} + \hat{j} + \hat{k})$.

$5 \vec{\beta}_2 = 5 \left(\frac{9}{5} \hat{i} + \frac{13}{5} \hat{j} - 3 \hat{k}\right) = 9 \hat{i} + 13 \hat{j} - 15 \hat{k}$ $(9 \hat{i} + 13 \hat{j} - 15 \hat{k}) \cdot (\hat{i} + \hat{j} + \hat{k}) = 9 \cdot 1 + 13 \cdot 1 + (-15) \cdot 1$ = 9 + 13 - 15 = 7\)

Therefore, the value of $5 \vec{\beta}_2 \cdot (\hat{i} + \hat{j} + \hat{k})$ is $\boxed{7}$.

Answer 2