Question

Find the equation of the line passing through the point A(1, 0, -1) and parallel to the
line joining B(1, 2, 3) and C( -1, 2, 0). Is it perpendicular to the line
(1 + 3α , 0 ,1 - 2α ).

Answer 1

Answer on Question #68959 – Math – Linear Algebra

Question

Find the equation of the line passing through the point $A(1, 0, -1)$ and parallel to the line joining $B(1, 2, 3)$ and $C(-1, 2, 0)$. Is it perpendicular to the line $(1 + 3\alpha, 0, 1 - 2\alpha)$?

Solution

1) Note that the line passing through $B(1, 2, 3)$ and $C(-1, 2, 0)$ has the direction vector $\vec{s} = \overrightarrow{BC} = (x_C - x_B, y_C - y_B, z_C - z_B) = (-1 - 1, 2 - 2, 0 - 3) = (-2, 0, -3)$;

2) The equation of the line passing through the point $A(x_0, y_0, z_0)$ and parallel to the line with direction vector $\vec{s} = (m, n, p)$ has the canonical form:

$$\frac{x - x_0}{m} = \frac{y - y_0}{n} = \frac{z - z_0}{p}$$

and the parametric form:

$$\left\{ \begin{array}{l} x = x_0 + m\alpha, \\ y = y_0 + n\alpha, \\ z = z_0 + p\alpha. \end{array} \right.$$

We have $A(1, 0, -1)$, $\vec{s} = (-2, 0, -3)$, substituting one gets

$$\frac{x - 1}{-2} = \frac{y - 0}{0} = \frac{z - (-1)}{-3};$$ $$l: \frac{x - 1}{-2} = \frac{y - 0}{0} = \frac{z + 1}{-3}.$$

3) The line $(1 + 3\alpha, 0, 1 - 2\alpha)$ has the next parametric form:

$$\left\{ \begin{array}{l} x = 1 + 3\alpha, \\ y = 0, \\ z = 1 - 2\alpha. \end{array} \right.$$

After reducing this form to the canonical equation

$$l_1: \frac{x - 1}{3} = \frac{y - 0}{0} = \frac{z - 1}{-2}.$$

We note that the line has the direction vector $\vec{s}_1 = (3, 0, -2)$.

4) The line $l$ with the direction vector $s$ is perpendicular to the line $l_1$ with the direction vector $s_1$ iff the scalar product is

$$s \cdot s_1 = 0.$$

Note that

$$s \cdot s_1 = m \cdot m_1 + n \cdot n_1 + p \cdot p_1 = -2 \cdot 3 + 0 \cdot 0 + (-3) \cdot (-2) = -6 + 6 = 0,$$

so line $l$ is perpendicular to the line $l_1$.

**Answer**: $l: \frac{x - 1}{-2} = \frac{y - 0}{0} = \frac{z + 1}{-3}$; it is perpendicular to the line $(1 + 3\alpha, 0, 1 - 2\alpha)$.

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