Let AB be the lines joining the points (1,-1,2) and (3,4,-2), and CD be the lines joining the points, (0,3,2), (3,5,6).
The direction ratios: a1,b1,c1 of AB are (3-1), (4-(-1)), and (-2-2) i.e., 2, 5, and -4.
The direction ratios: a2,b2,c2 of CD are (3-0), (5-3), and (6-2) i.e., 3, 2, and 4.
AB and CD will be perpendicular to each other if a1a2+b1b2+c1c2=0
a1a2+b1b2+c1c2
=2×3+5×2+(-4)×4
=6+10-16
=0
Therefore, AB and CD are perpendicular to each other.
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:
\[ \frac{x - 8}{3} = \frac{y + 16}{7} = \frac{z - 10}{-16} \quad \text{and} \quad \frac{x - 15}{3} = \frac{y - 29}{-8} = \frac{z - 5}{-5} \]
Analyse the characters of William Douglas from ‘Deep Water’ and Mukesh from ‘Lost Spring’ in terms of their determination and will power in pursuing their goals.
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\(\hat{k}\)