Question

Let \(\vec{p}=2\hat{i}+\hat{j}+3\hat{k}\) and \(\vec{q}=\hat{i}-\hat{j}+\hat{k}\). If for some real numbers α, β and γ we have
\(15\hat{i}+10\hat{j}+6\hat{k}=α(2\vec{p}+\vec{q})+β(\vec{p}-2\vec{q})+γ(\vec{p}\times\vec{q})\),
then the value of γ is ________.

Answer 1

Correct Answer: 2

To find \(\gamma\), express the given vector in terms of \(\vec{p}\), \(\vec{q}\), and their cross product, then solve for the coefficients.

Given:
\[ \vec{p} = 2\hat{i} + \hat{j} + 3\hat{k}, \quad \vec{q} = \hat{i} - \hat{j} + \hat{k} \] \[ \vec{v} = 15\hat{i} + 10\hat{j} + 6\hat{k} = \alpha (2\vec{p} + \vec{q}) + \beta (\vec{p} - 2\vec{q}) + \gamma (\vec{p} \times \vec{q}) \]

Step 1: Calculate \(2\vec{p} + \vec{q}\) and \(\vec{p} - 2\vec{q}\)
\[ 2\vec{p} = 4\hat{i} + 2\hat{j} + 6\hat{k} \] \[ 2\vec{p} + \vec{q} = (4+1)\hat{i} + (2 - 1)\hat{j} + (6 + 1)\hat{k} = 5\hat{i} + \hat{j} + 7\hat{k} \] \[ 2\vec{q} = 2\hat{i} - 2\hat{j} + 2\hat{k} \] \[ \vec{p} - 2\vec{q} = (2 - 2)\hat{i} + (1 + 2)\hat{j} + (3 - 2)\hat{k} = 0\hat{i} + 3\hat{j} + 1\hat{k} \]

Step 2: Calculate \(\vec{p} \times \vec{q}\)
\[ \vec{p} \times \vec{q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 3 \\ 1 & -1 & 1 \end{vmatrix} = \hat{i} (1 \times 1 - 3 \times (-1)) - \hat{j} (2 \times 1 - 3 \times 1) + \hat{k} (2 \times (-1) - 1 \times 1) \] \[ = \hat{i} (1 + 3) - \hat{j} (2 - 3) + \hat{k} (-2 - 1) = 4 \hat{i} + 1 \hat{j} - 3 \hat{k} \]

Step 3: Write the equation for \(\vec{v}\) components:
\[ \vec{v} = \alpha (5\hat{i} + \hat{j} + 7\hat{k}) + \beta (0\hat{i} + 3\hat{j} + \hat{k}) + \gamma (4\hat{i} + \hat{j} - 3\hat{k}) \] \[ = (5\alpha + 0\beta + 4\gamma) \hat{i} + (\alpha + 3\beta + \gamma) \hat{j} + (7\alpha + \beta - 3\gamma) \hat{k} \] Set equal to \(\vec{v} = 15\hat{i} + 10\hat{j} + 6\hat{k}\): \[ 5\alpha + 4\gamma = 15 \quad (1) \] \[ \alpha + 3\beta + \gamma = 10 \quad (2) \] \[ 7\alpha + \beta - 3\gamma = 6 \quad (3) \]

Step 4: Solve equations (1), (2), and (3)
From (1): \[ 5\alpha = 15 - 4\gamma \implies \alpha = \frac{15 - 4\gamma}{5} = 3 - \frac{4}{5} \gamma \] Substitute \(\alpha\) into (2): \[ 3 - \frac{4}{5} \gamma + 3\beta + \gamma = 10 \] \[ 3\beta + \left(3 - \frac{4}{5}\gamma + \gamma \right) = 10 \] \[ 3\beta + 3 + \frac{1}{5} \gamma = 10 \] \[ 3\beta = 7 - \frac{1}{5} \gamma \] \[ \beta = \frac{7}{3} - \frac{\gamma}{15} \] Substitute \(\alpha\) and \(\beta\) into (3): \[ 7\left(3 - \frac{4}{5} \gamma \right) + \left( \frac{7}{3} - \frac{\gamma}{15} \right) - 3 \gamma = 6 \] \[ 21 - \frac{28}{5} \gamma + \frac{7}{3} - \frac{\gamma}{15} - 3 \gamma = 6 \] Combine constants: \[ 21 + \frac{7}{3} = \frac{63}{3} + \frac{7}{3} = \frac{70}{3} \] Combine \(\gamma\) terms (common denominator 15): \[ - \frac{28}{5} \gamma - \frac{\gamma}{15} - 3 \gamma = - \frac{84}{15} \gamma - \frac{1}{15} \gamma - \frac{45}{15} \gamma = - \frac{130}{15} \gamma = - \frac{26}{3} \gamma \] Equation becomes: \[ \frac{70}{3} - \frac{26}{3} \gamma = 6 \] Multiply both sides by 3: \[ 70 - 26 \gamma = 18 \] \[ 70 - 18 = 26 \gamma \] \[ 52 = 26 \gamma \] \[ \gamma = 2 \]

Final Answer:
\[ \boxed{2} \]

Answer 2

Vector Operations 

We are given:

\[ \vec{p} = 2\hat{i} + \hat{j} + 3\hat{k}, \quad \vec{q} = \hat{i} - \hat{j} + \hat{k} \]

Step 1: Calculate Linear Combinations of Vectors

First, compute:

\[ 2\vec{p} + \vec{q} = 2(2\hat{i} + \hat{j} + 3\hat{k}) + (\hat{i} - \hat{j} + \hat{k}) \]

Simplify:

\[ 2\vec{p} + \vec{q} = 4\hat{i} + 2\hat{j} + 6\hat{k} + \hat{i} - \hat{j} + \hat{k} = 5\hat{i} + \hat{j} + 7\hat{k} \]

Next, compute:

\[ \vec{p} - 2\vec{q} = (2\hat{i} + \hat{j} + 3\hat{k}) - 2(\hat{i} - \hat{j} + \hat{k}) \]

Simplify:

\[ \vec{p} - 2\vec{q} = 2\hat{i} + \hat{j} + 3\hat{k} - 2\hat{i} + 2\hat{j} - 2\hat{k} = \hat{j} + \hat{k} \]

Step 2: Compute the Cross Product \(\vec{p} \times \vec{q}\)

Using the determinant formula for the cross product:

\[ \vec{p} \times \vec{q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 3 \\ 1 & -1 & 1 \end{vmatrix} \]

Expanding the determinant:

\[ \vec{p} \times \vec{q} = \hat{i} \begin{vmatrix} 1 & 3 \\ -1 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 3 \\ 1 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & 1 \\ 1 & -1 \end{vmatrix} \]

Simplify:

\[ \vec{p} \times \vec{q} = \hat{i}(1 \cdot 1 - (-1 \cdot 3)) - \hat{j}(2 \cdot 1 - 3 \cdot 1) + \hat{k}(2 \cdot (-1) - 1 \cdot 1) \] \[ \vec{p} \times \vec{q} = \hat{i}(1 + 3) - \hat{j}(2 - 3) + \hat{k}(-2 - 1) \] \[ \vec{p} \times \vec{q} = 4\hat{i} + \hat{j} - 3\hat{k} \]

Step 3: Form and Solve the Equation

We now solve for \(\alpha\), \(\beta\), and \(\gamma\) such that:

\[ \alpha(5\hat{i} + \hat{j} + 7\hat{k}) + \beta(\hat{j} + \hat{k}) + \gamma(4\hat{i} + \hat{j} - 3\hat{k}) = 15\hat{i} + 10\hat{j} + 6\hat{k} \]

Equating Components:

- For \(\hat{i}\): \[ 5\alpha + 4\gamma = 15 \] - For \(\hat{j}\): \[ \alpha + \beta + \gamma = 10 \] - For \(\hat{k}\): \[ 7\alpha + \beta - 3\gamma = 6 \]

Step 4: Solve the System of Equations

From the first equation:

\[ \alpha = \frac{15 - 4\gamma}{5} \]

Substitute \(\alpha\) into the other two equations and solve:

\[ \beta = \frac{11}{5}, \quad \gamma = 2, \quad \alpha = \frac{7}{5} \]

Final Answer:

• \(\alpha = \frac{7}{5}, \beta = \frac{11}{5}, \gamma = 2\)