Question

Let O be the origin and OA = a1i +a2j + a3k. find the equation of a line that passes through the point (1 2 3) and is perpendicular to the vector 3i+2j+3k.

Answer 1

Answer on Question #63631 – Math – Analytic Geometry

Question

Let $O$ be the origin and $OA = a1i + a2j + a3k$. Find the equation of a line that passes through the point $(1\ 2\ 3)$ and is perpendicular to the vector $3i + 2j + 3k$.

Solution

There are two ways of solution of this problem

1. The first method

$\vec{a} = m(3\vec{i} + 2\vec{j} + 3\vec{k})$ is a projection of $\vec{b}$ on $\vec{c}$

$$\vec {b} = \vec {i} + 2 \vec {j} + 3 \vec {k}$$ $$\vec {b} + \vec {x} = \vec {a}, \quad \vec {x} = \vec {a} - \vec {b}$$ $$\vec {x} = (3 m - 1) \vec {i} + (2 m - 2) \vec {j} + (3 m - 3) \vec {k}$$

Since $\vec{a} \perp \vec{x}$ then $\vec{a} \cdot \vec{x} = 0$ or

$$\left((3 m - 1) \vec {i} + (2 m - 2) \vec {j} + (3 m - 3) \vec {k}\right) \cdot \left(3 \vec {i} + 2 \vec {j} + 3 \vec {k}\right) = 0$$ $$9 m - 3 + 4 m - 4 + 9 m - 9 = 0$$ $$2 2 m = 1 6$$ $$m = \frac {8}{1 1}$$ $$\vec {x} = \left(\frac {2 4}{1 1} - 1\right) \vec {i} + \left(\frac {1 6}{1 1} - 2\right) \vec {j} + \left(\frac {2 4}{1 1} - 3\right) \vec {k}$$

Thus, the direction vector of the line is

$$\vec{n} = \frac{13}{11} \vec{i} - \frac{6}{11} \vec{j} - \frac{9}{11} \vec{k}.$$

One more collinear vector is $11\left(\frac{13}{11}, - \frac{6}{11}, - \frac{9}{11}\right) = (13, - 6, - 9).$

The line passes through the point (1 2 3).

Thus, the equation of the line is

$$\frac{x - 1}{13} = \frac{y - 2}{-6} = \frac{z - 3}{-9}$$

Answer: $\frac{x - 1}{13} = \frac{y - 2}{-6} = \frac{z - 3}{-9}$

2. The second method

$\vec{n}$ is a normal to the plane, where the vectors $\vec{a}$ and $\vec{b}$ lie,

$$\begin{aligned} \vec{n} = \vec{a} \times \vec{b} &= \left| \begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \\ 3 & 2 & 3 \\ 1 & 2 & 3 \end{array} \right| = (2 \cdot 3 - 2 \cdot 3) \vec{i} - (3 \cdot 3 - 1 \cdot 3) \vec{j} + (3 \cdot 2 - 1 \cdot 2) \vec{k} = \\ &= (6 - 6) \vec{i} - (9 - 3) \vec{j} + (6 - 2) \vec{k} = 0 \vec{i} - 6 \vec{j} + 4 \vec{k} \end{aligned}$$

$a \times n$ is a vector that lies on this plane and is perpendicular to $\vec{a}$, and it is the direction vector of the line

$$a \times n = \left| \begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \\ 3 & 2 & 3 \\ 0 & -6 & 4 \end{array} \right| = (8 + 18) \vec{i} - (12 - 0) \vec{j} + (-18 - 0) \vec{k} = 26 \vec{i} - 12 \vec{j} - 18 \vec{k}$$

This is the direction vector of the line and one more collinear one is

$$\left(\frac{26}{2}, - \frac{12}{2}, - \frac{18}{2}\right) = (13, - 6, - 9).$$

The line passes through the point (1 2 3). Using the condition for collinearity of vectors we get that the equation of the line is

$$\frac{x - 1}{13} = \frac{y - 2}{-6} = \frac{z - 3}{-9}$$

Answer: $\frac{x - 1}{13} = \frac{y - 2}{-6} = \frac{z - 3}{-9}$

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