Question

Determine the vector equation of the line that passes through the point \( (1, 2, -3) \) and is perpendicular to both of the following lines:

\[ \frac{x - 8}{3} = \frac{y + 16}{7} = \frac{z - 10}{-16} \quad \text{and} \quad \frac{x - 15}{3} = \frac{y - 29}{-8} = \frac{z - 5}{-5} \]

Answer 1

Find the Vector Equation of a Line

Given:

We are to find the vector equation of a line passing through the point \( (1, 2, -3) \) and perpendicular to both the lines:

Line 1: \[ \frac{x - 8}{3} = \frac{y + 16}{7} = \frac{z - 10}{-16} \] Line 2: \[ \frac{x - 15}{3} = \frac{y - 29}{-8} = \frac{z - 5}{-5} \]

Step 1: Extract Direction Vectors

Direction vector of Line 1: \[ \vec{d}_1 = \langle 3, 7, -16 \rangle \] Direction vector of Line 2: \[ \vec{d}_2 = \langle 3, -8, -5 \rangle \]

Step 2: Find Cross Product

The required line is perpendicular to both lines, so we compute the cross product:

\[ \vec{n} = \vec{d}_1 \times \vec{d}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 7 & -16 \\ 3 & -8 & -5 \end{vmatrix} \] \[ = \hat{i}(7 \cdot (-5) - (-16) \cdot (-8)) - \hat{j}(3 \cdot (-5) - (-16) \cdot 3) + \hat{k}(3 \cdot (-8) - 7 \cdot 3) \] \[ = \hat{i}(-35 - 128) - \hat{j}(-15 + 48) + \hat{k}(-24 - 21) \] \[ = \hat{i}(-163) - \hat{j}(33) + \hat{k}(-45) \Rightarrow \vec{n} = \langle -163, -33, -45 \rangle \]

Step 3: Use Point (1, 2, -3)

Line passes through point \( \vec{a} = \langle 1, 2, -3 \rangle \)

Step 4: Vector Equation of the Required Line

\[ \vec{r} = \vec{a} + \lambda \vec{n} = \langle 1, 2, -3 \rangle + \lambda \langle -163, -33, -45 \rangle \]

Step 5: Parametric Form

\[ x = 1 - 163\lambda,\quad y = 2 - 33\lambda,\quad z = -3 - 45\lambda \]

✅ Final Answer:

The vector equation is: \[ \boxed{ \vec{r} = \langle 1, 2, -3 \rangle + \lambda \langle -163, -33, -45 \rangle } \]