Let AB be the line through the points (4,7,8), and (2,3,4), and CD be the line through the points (-1,-2,1), and (1,2,5).
The direction ratios, a1, b1, c1 of AC are (2-4), (3-7), and (4-8) i.e., -2, -4, and -4.
The direction ratios, a2, b2, c2 of CD are (1-(-1)), (2-(-2)), and (5-1) i.e., 2, 4, and 4.
AB will parellel to CD, if \(\frac{a_1}{a_2}\)=\(\frac{b_1}{b_2}\)=\(\frac{c_1}{c_2}\)
\(\frac{a_1}{a_2}\)
=\(-\frac{2}{2}\)
=-1
\(\frac{b_1}{b_2}\)
=\(-\frac{4}{4}\)
=-1
\(\frac{c_1}{c_2}\)
=\(-\frac{4}{4}\)
=-1
∴ \(\frac{a_1}{a_2}\) = \(\frac{b_1}{b_2}\) = \(\frac{c_1}{c_2}\)
Thus, AB is parallel to CD.
Show that the following lines intersect. Also, find their point of intersection:
Line 1: \[ \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} \]
Line 2: \[ \frac{x - 4}{5} = \frac{y - 1}{2} = z \]
The vector equations of two lines are given as:
Line 1: \[ \vec{r}_1 = \hat{i} + 2\hat{j} - 4\hat{k} + \lambda(4\hat{i} + 6\hat{j} + 12\hat{k}) \]
Line 2: \[ \vec{r}_2 = 3\hat{i} + 3\hat{j} - 5\hat{k} + \mu(6\hat{i} + 9\hat{j} + 18\hat{k}) \]
Determine whether the lines are parallel, intersecting, skew, or coincident. If they are not coincident, find the shortest distance between them.
Determine the vector equation of the line that passes through the point \( (1, 2, -3) \) and is perpendicular to both of the following lines:
\[ \frac{x - 8}{3} = \frac{y + 16}{7} = \frac{z - 10}{-16} \quad \text{and} \quad \frac{x - 15}{3} = \frac{y - 29}{-8} = \frac{z - 5}{-5} \]
In a plane, the equation of a line is given by the popular equation y = m x + C. Let's look at how the equation of a line is written in vector form and Cartesian form.
Consider a line that passes through a given point, say ‘A’, and the line is parallel to a given vector '\(\vec{b}\)‘. Here, the line ’l' is given to pass through ‘A’, whose position vector is given by '\(\vec{a}\)‘. Now, consider another arbitrary point ’P' on the given line, where the position vector of 'P' is given by '\(\vec{r}\)'.
\(\vec{AP}\)=𝜆\(\vec{b}\)
Also, we can write vector AP in the following manner:
\(\vec{AP}\)=\(\vec{OP}\)–\(\vec{OA}\)
𝜆\(\vec{b}\) =\(\vec{r}\)–\(\vec{a}\)
\(\vec{a}\)=\(\vec{a}\)+𝜆\(\vec{b}\)
\(\vec{b}\)=𝑏1\(\hat{i}\)+𝑏2\(\hat{j}\) +𝑏3\(\hat{k}\)