If `veca=3hati+4hatj+5hatk` and `vecbeta =2hati+hatj-4hatk` , then express `vecbeta=vecbeta_1+vecbeta_2` such that `vecbeta_1||vecalpha` and `vecbeta_2botvecalpha`.

Answer

To solve the problem, we need to express the vector \(\vec{\beta}\) as the sum of two vectors \(\vec{\beta_1}\) and \(\vec{\beta_2}\), where \(\vec{\beta_1}\) is parallel to \(\vec{\alpha}\) and \(\vec{\beta_2}\) is perpendicular to \(\vec{\alpha}\).

Given:
\[ \vec{\alpha} = 3\hat{i} + 4\hat{j} + 5\hat{k} \]
\[ \vec{\beta} = 2\hat{i} + \hat{j} - 4\hat{k} \]

### Step 1: Express \(\vec{\beta_1}\) in terms of \(\vec{\alpha}\)

Since \(\vec{\beta_1}\) is parallel to \(\vec{\alpha}\), we can express it as:
\[ \vec{\beta_1} = \lambda \vec{\alpha} \]
where \(\lambda\) is a scalar.

Substituting \(\vec{\alpha}\):
\[ \vec{\beta_1} = \lambda(3\hat{i} + 4\hat{j} + 5\hat{k}) = 3\lambda \hat{i} + 4\lambda \hat{j} + 5\lambda \hat{k} \]

### Step 2: Express \(\vec{\beta_2}\)

We know that:
\[ \vec{\beta} = \vec{\beta_1} + \vec{\beta_2} \]
Thus, we can express \(\vec{\beta_2}\) as:
\[ \vec{\beta_2} = \vec{\beta} - \vec{\beta_1} \]
Substituting the expressions for \(\vec{\beta}\) and \(\vec{\beta_1}\):
\[ \vec{\beta_2} = (2\hat{i} + \hat{j} - 4\hat{k}) - (3\lambda \hat{i} + 4\lambda \hat{j} + 5\lambda \hat{k}) \]
This simplifies to:
\[ \vec{\beta_2} = (2 - 3\lambda)\hat{i} + (1 - 4\lambda)\hat{j} + (-4 - 5\lambda)\hat{k} \]

### Step 3: Use the condition that \(\vec{\beta_2}\) is perpendicular to \(\vec{\alpha}\)

For \(\vec{\beta_2}\) to be perpendicular to \(\vec{\alpha}\), their dot product must equal zero:
\[ \vec{\beta_2} \cdot \vec{\alpha} = 0 \]
Substituting the expressions:
\[ [(2 - 3\lambda)\hat{i} + (1 - 4\lambda)\hat{j} + (-4 - 5\lambda)\hat{k}] \cdot (3\hat{i} + 4\hat{j} + 5\hat{k}) = 0 \]
Calculating the dot product:
\[ (2 - 3\lambda) \cdot 3 + (1 - 4\lambda) \cdot 4 + (-4 - 5\lambda) \cdot 5 = 0 \]
Expanding:
\[ 6 - 9\lambda + 4 - 16\lambda - 20 - 25\lambda = 0 \]
Combining like terms:
\[ (6 + 4 - 20) + (-9\lambda - 16\lambda - 25\lambda) = 0 \]
This simplifies to:
\[ -10 - 50\lambda = 0 \]
Solving for \(\lambda\):
\[ 50\lambda = -10 \implies \lambda = -\frac{1}{5} \]

### Step 4: Substitute \(\lambda\) back to find \(\vec{\beta_1}\) and \(\vec{\beta_2}\)

Substituting \(\lambda\) into \(\vec{\beta_1}\):
\[ \vec{\beta_1} = 3\left(-\frac{1}{5}\right)\hat{i} + 4\left(-\frac{1}{5}\right)\hat{j} + 5\left(-\frac{1}{5}\right)\hat{k} \]
Calculating:
\[ \vec{\beta_1} = -\frac{3}{5}\hat{i} - \frac{4}{5}\hat{j} - 1\hat{k} \]

Now substituting \(\lambda\) into \(\vec{\beta_2}\):
\[ \vec{\beta_2} = (2 - 3(-\frac{1}{5}))\hat{i} + (1 - 4(-\frac{1}{5}))\hat{j} + (-4 - 5(-\frac{1}{5}))\hat{k} \]
Calculating:
\[ \vec{\beta_2} = \left(2 + \frac{3}{5}\right)\hat{i} + \left(1 + \frac{4}{5}\right)\hat{j} + \left(-4 + 1\right)\hat{k} \]
This simplifies to:
\[ \vec{\beta_2} = \frac{10}{5} + \frac{3}{5} = \frac{13}{5}\hat{i} + \frac{9}{5}\hat{j} - 3\hat{k} \]

### Final Result

Thus, we have:
\[ \vec{\beta_1} = -\frac{3}{5}\hat{i} - \frac{4}{5}\hat{j} - 1\hat{k} \]
\[ \vec{\beta_2} = \frac{13}{5}\hat{i} + \frac{9}{5}\hat{j} - 3\hat{k} \]