7.5 Q-4

Question Statement

We are tasked with finding the value of $\alpha$ such that the vectors $\mathbf{a}$, $\mathbf{b}$, and $\mathbf{c}$ are coplanar.

Part (i)

• $\mathbf{a} = \hat{i} - \hat{j} + \hat{k}$
• $\mathbf{b} = \hat{i} - 2\hat{j} - 3\hat{k}$
• $\mathbf{c} = 3\hat{i} - \alpha \hat{j} + 5\hat{k}$

Part (ii)

• $\mathbf{a} = \hat{i} - 2\alpha \hat{j} + \hat{k}$
• $\mathbf{b} = \hat{i} - \hat{j} + 2\hat{k}$
• $\mathbf{c} = \alpha \hat{i} - \hat{j} + \hat{k}$

We need to find $\alpha$ such that these vectors are coplanar.

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Background and Explanation

Vectors are coplanar if the scalar triple product of the three vectors equals zero. The scalar triple product is given by:

$$\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = \left| \begin{array}{ccc} a_1 & a_2 & a_3 b_1 & b_2 & b_3 c_1 & c_2 & c_3 \end{array} \right|$$

If the determinant of this 3x3 matrix (the scalar triple product) is zero, the vectors are coplanar.

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Solution

Part (i)

Given the vectors:

$$\mathbf{a} = \hat{i} - \hat{j} + \hat{k}, \quad \mathbf{b} = \hat{i} - 2\hat{j} - 3\hat{k}, \quad \mathbf{c} = 3\hat{i} - \alpha \hat{j} + 5\hat{k}$$

We form the determinant for the scalar triple product:

$$\left| \begin{array}{ccc} 1 & -1 & 1 1 & -2 & -3 3 & -\alpha & 5 \end{array} \right| = 0$$

We now expand the determinant along the first row:

$$= 1 \left( \begin{vmatrix} -2 & -3 -\alpha & 5 \end{vmatrix} \right) - (-1) \left( \begin{vmatrix} 1 & -3 3 & 5 \end{vmatrix} \right) + 1 \left( \begin{vmatrix} 1 & -2 3 & -\alpha \end{vmatrix} \right)$$

Now, we compute the 2x2 determinants:

$$\begin{vmatrix} -2 & -3 -\alpha & 5 \end{vmatrix} = (-2)(5) - (-3)(-\alpha) = -10 - 3\alpha$$ $$\begin{vmatrix} 1 & -3 3 & 5 \end{vmatrix} = (1)(5) - (-3)(3) = 5 + 9 = 14$$ $$\begin{vmatrix} 1 & -2 3 & -\alpha \end{vmatrix} = (1)(-\alpha) - (-2)(3) = -\alpha + 6$$

Substitute these into the determinant expression:

$$= 1(-10 - 3\alpha) + 1(14) + 1(-\alpha + 6)$$ $$= -10 - 3\alpha + 14 - \alpha + 6$$ $$= 10 - 4\alpha$$

For the vectors to be coplanar, the determinant must be zero:

$$10 - 4\alpha = 0$$

Solving for $\alpha$:

$$4\alpha = 10 \quad \Rightarrow \quad \alpha = \frac{10}{4} = 2.5$$

Part (ii)

Given the vectors:

$$\mathbf{a} = \hat{i} - 2\alpha \hat{j} + \hat{k}, \quad \mathbf{b} = \hat{i} - \hat{j} + 2\hat{k}, \quad \mathbf{c} = \alpha \hat{i} - \hat{j} + \hat{k}$$

We form the determinant for the scalar triple product:

$$\left| \begin{array}{ccc} 1 & -2\alpha & -1 1 & -1 & 2 \alpha & -1 & 1 \end{array} \right| = 0$$

We now expand the determinant along the first row:

$$= 1 \left( \begin{vmatrix} -1 & 2 -1 & 1 \end{vmatrix} \right) - (-2\alpha) \left( \begin{vmatrix} 1 & 2 \alpha & 1 \end{vmatrix} \right) + (-1) \left( \begin{vmatrix} 1 & -1 \alpha & -1 \end{vmatrix} \right)$$

Now, we compute the 2x2 determinants:

$$\begin{vmatrix} -1 & 2 -1 & 1 \end{vmatrix} = (-1)(1) - (2)(-1) = -1 + 2 = 1$$ $$\begin{vmatrix} 1 & 2 \alpha & 1 \end{vmatrix} = (1)(1) - (2)(\alpha) = 1 - 2\alpha$$ $$\begin{vmatrix} 1 & -1 \alpha & -1 \end{vmatrix} = (1)(-1) - (-1)(\alpha) = -1 + \alpha$$

Substitute these into the determinant expression:

$$= 1(1) - (-2\alpha)(1 - 2\alpha) - 1(-1 + \alpha)$$ $$= 1 + 2\alpha(1 - 2\alpha) + (1 - \alpha)$$ $$= 1 + 2\alpha - 4\alpha^2 + 1 - \alpha$$ $$= 2 + \alpha - 4\alpha^2$$

For the vectors to be coplanar, the determinant must be zero:

$$2 + \alpha - 4\alpha^2 = 0$$

Rearranging into a standard quadratic form:

$$4\alpha^2 - \alpha - 2 = 0$$

We solve this quadratic equation using the quadratic formula:

$$\alpha = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(4)(-2)}}{2(4)}$$ $$\alpha = \frac{1 \pm \sqrt{1 + 32}}{8}$$ $$\alpha = \frac{1 \pm \sqrt{33}}{8}$$

Thus, the two possible values for $\alpha$ are:

$$\alpha = \frac{1 + \sqrt{33}}{8} \quad \text{or} \quad \alpha = \frac{1 - \sqrt{33}}{8}$$

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Key Formulas or Methods Used

• Scalar Triple Product:

$$\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = \left| \begin{array}{ccc} a_1 & a_2 & a_3 b_1 & b_2 & b_3 c_1 & c_2 & c_3 \end{array} \right|$$

• Quadratic Formula for solving quadratic equations.

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Summary of Steps

1. For part (i), set up the determinant for the scalar triple product and expand it.
2. Solve the resulting equation to find $\alpha = 2.5$.
3. For part (ii), form and expand the determinant for the scalar triple product.
4. Solve the resulting quadratic equation using the quadratic formula to find two possible values for $\alpha$: $\frac{1 + \sqrt{33}}{8}$ and $\frac{1 - \sqrt{33}}{8}$.

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