7.5 Q-3

Question Statement

We are asked to prove that the vectors:

$$\mathbf{a} = \hat{i} - 2 \hat{j} + 3 \hat{k}, \quad \mathbf{b} = -2 \hat{i} + 3 \hat{j} - 4 \hat{k}, \quad \mathbf{c} = \hat{i} - 3 \hat{j} + 5 \hat{k}$$

are coplanar. To do this, we need to show that the scalar triple product of these vectors is zero, as this indicates that the vectors lie in the same plane.

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Background and Explanation

For three vectors to be coplanar, the scalar triple product must be zero. The scalar triple product of three vectors $\mathbf{a}$, $\mathbf{b}$, and $\mathbf{c}$ is given by:

$$\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})$$

This can also be represented as the determinant of a 3x3 matrix formed by the components of the vectors:

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

If the scalar triple product is zero, the vectors are coplanar. If it’s non-zero, the vectors are not coplanar.

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Solution

We are given the following vectors:

$$\mathbf{a} = \hat{i} - 2 \hat{j} + 3 \hat{k}, \quad \mathbf{b} = -2 \hat{i} + 3 \hat{j} - 4 \hat{k}, \quad \mathbf{c} = \hat{i} - 3 \hat{j} + 5 \hat{k}$$

To prove that these vectors are coplanar, we need to calculate the scalar triple product $\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})$.

Step 1: Set up the determinant for the scalar triple product

The scalar triple product is given by:

$$\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = \left| \begin{array}{ccc} 1 & -2 & 3 -2 & 3 & -4 -3 & -3 & 5 \end{array} \right|$$

Where the rows are the components of $\mathbf{a}$, $\mathbf{b}$, and $\mathbf{c}$, respectively.

Step 2: Expand the determinant

We now expand the determinant along the first row:

$$\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = 1 \cdot \begin{vmatrix} 3 & -4 -3 & 5 \end{vmatrix} - (-2) \cdot \begin{vmatrix} -2 & -4 -3 & 5 \end{vmatrix} + 3 \cdot \begin{vmatrix} -2 & 3 -3 & -3 \end{vmatrix}$$

Step 3: Compute the 2x2 determinants

Now, compute each of the 2x2 determinants:

$$\begin{vmatrix} 3 & -4 -3 & 5 \end{vmatrix} = (3)(5) - (-4)(-3) = 15 - 12 = 3$$ $$\begin{vmatrix} -2 & -4 -3 & 5 \end{vmatrix} = (-2)(5) - (-4)(-3) = -10 - 12 = -22$$ $$\begin{vmatrix} -2 & 3 -3 & -3 \end{vmatrix} = (-2)(-3) - (3)(-3) = 6 + 9 = 15$$

Step 4: Substitute the results into the determinant

Now substitute these values back into the original expansion:

$$\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = 1(3) - (-2)(-22) + 3(15)$$ $$= 3 - 44 + 45$$ $$= 3 - 44 + 45 = 12 - 12 = 0$$

Step 5: Conclusion

Since the scalar triple product is zero, we can conclude that the vectors $\mathbf{a}$, $\mathbf{b}$, and $\mathbf{c}$ are coplanar.

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Key Formulas or Methods Used

• Scalar Triple Product:

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

• The property that if the scalar triple product is zero, the vectors are coplanar.

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Summary of Steps

1. Write down the given vectors $\mathbf{a}$, $\mathbf{b}$, and $\mathbf{c}$.
2. Set up the determinant for the scalar triple product.
3. Expand the determinant along the first row.
4. Calculate the 2x2 determinants.
5. Substitute the values into the determinant expression.
6. Verify that the scalar triple product equals zero, proving that the vectors are coplanar.

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