Advertisements
Advertisements
प्रश्न
Find the vector and Cartesian equations of the line through the point (1, 2, −4) and perpendicular to the two lines.
`vecr=(8hati-19hatj+10hatk)+lambda(3hati-16hatj+7hatk) " and "vecr=(15hati+29hatj+5hatk)+mu(3hati+8hatj-5hatk)`
Advertisements
उत्तर
The equations of the given lines are
`vecr=(8hati-19hatj+10hatk)+lambda(3hati-16hatj+7hatk) " ...(1)"`
`vecr=(15hati+29hatj+5hatk)+mu(3hati+8hatj-5hatk) " .....(2)"`
Normal parallel to (1) is `vec(n_1)=3hati-16hatj+7hatk`
Normal parallel to (2) is `vec(n_2)=3hati+8hatj-5hatk`
The required line is perpendicular to the given lines. So, the normal n⃗ parallel to the required line is perpendicular to `vec(n_1) " and "vec(n_2)`
`:.vecn=vec(n_1)xxvec(n_2)=|(hati, hatj,hatk),(3,-16,7),(3,8,-5)|=24hati+36hatj+72hatk`
Thus, the vector equation of the required line is
`vecr=(hati+2hatj-4hatk)+gamma(24hati+36hatj+72hatk)`
`=>vecr=(hati+2hatj-4hatk)+k(2hati+3hatj+6hatk) " (Where k=12γ)"`
Also, the Cartesian equation of the required line is
`(x-1)/2=(y-2)/3=(z+4)/6`
APPEARS IN
संबंधित प्रश्न
Find the value of p, so that the lines `l_1:(1-x)/3=(7y-14)/p=(z-3)/2 and l_2=(7-7x)/3p=(y-5)/1=(6-z)/5 ` are perpendicular to each other. Also find the equations of a line passing through a point (3, 2, – 4) and parallel to line l1.
Find the vector and the Cartesian equations of the line that passes through the points (3, −2, −5), (3, −2, 6).
Find the equation of a line parallel to x-axis and passing through the origin.
Find the vector equation of a line passing through (2, −1, 1) and parallel to the line whose equations are \[\frac{x - 3}{2} = \frac{y + 1}{7} = \frac{z - 2}{- 3} .\]
Find the vector equation of the line passing through the point A(1, 2, –1) and parallel to the line 5x – 25 = 14 – 7y = 35z.
Show that the line through the points (1, −1, 2) and (3, 4, −2) is perpendicular to the through the points (0, 3, 2) and (3, 5, 6).
Find the angle between the following pair of line:
\[\overrightarrow{r} = \left( 4 \hat{i} - \hat{j} \right) + \lambda\left( \hat{i} + 2 \hat{j} - 2 \hat{k} \right) \text{ and }\overrightarrow{r} = \hat{i} - \hat{j} + 2 \hat{k} - \mu\left( 2 \hat{i} + 4 \hat{j} - 4 \hat{k} \right)\]
Find the angle between the pairs of lines with direction ratios proportional to 2, 2, 1 and 4, 1, 8 .
Find the equations of the line passing through the point (2, 1, 3) and perpendicular to the lines \[\frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z - 3}{3} \text{ and } \frac{x}{- 3} = \frac{y}{2} = \frac{z}{5}\]
Find the equation of the line passing through the point \[\hat{i} + \hat{j} - 3 \hat{k} \] and perpendicular to the lines \[\overrightarrow{r} = \hat{i} + \lambda\left( 2 \hat{i} + \hat{j} - 3 \hat{k} \right) \text { and } \overrightarrow{r} = \left( 2 \hat{i} + \hat{j} - \hat{ k} \right) + \mu\left( \hat{i} + \hat{j} + \hat{k} \right) .\]
Show that the lines \[\frac{x}{1} = \frac{y - 2}{2} = \frac{z + 3}{3} \text{ and } \frac{x - 2}{2} = \frac{y - 6}{3} = \frac{z - 3}{4}\] intersect and find their point of intersection.
Prove that the lines through A (0, −1, −1) and B (4, 5, 1) intersects the line through C (3, 9, 4) and D (−4, 4, 4). Also, find their point of intersection.
Determine whether the following pair of lines intersect or not:
\[\frac{x - 1}{3} = \frac{y - 1}{- 1} = \frac{z + 1}{0} and \frac{x - 4}{2} = \frac{y - 0}{0} = \frac{z + 1}{3}\]
Find the perpendicular distance of the point (3, −1, 11) from the line \[\frac{x}{2} = \frac{y - 2}{- 3} = \frac{z - 3}{4} .\]
Find the foot of the perpendicular drawn from the point A (1, 0, 3) to the joint of the points B (4, 7, 1) and C (3, 5, 3).
Find the foot of perpendicular from the point (2, 3, 4) to the line \[\frac{4 - x}{2} = \frac{y}{6} = \frac{1 - z}{3} .\] Also, find the perpendicular distance from the given point to the line.
Find the equation of the perpendicular drawn from the point P (2, 4, −1) to the line \[\frac{x + 5}{1} = \frac{y + 3}{4} = \frac{z - 6}{- 9} .\] Also, write down the coordinates of the foot of the perpendicular from P.
Find the foot of the perpendicular from (0, 2, 7) on the line \[\frac{x + 2}{- 1} = \frac{y - 1}{3} = \frac{z - 3}{- 2} .\]
Find the shortest distance between the following pairs of lines whose cartesian equations are : \[\frac{x - 1}{2} = \frac{y + 1}{3} = z \text{ and } \frac{x + 1}{3} = \frac{y - 2}{1}; z = 2\]
Find the shortest distance between the following pairs of lines whose cartesian equations are : \[\frac{x - 1}{- 1} = \frac{y + 2}{1} = \frac{z - 3}{- 2} \text{ and } \frac{x - 1}{1} = \frac{y + 1}{2} = \frac{z + 1}{- 2}\]
Find the distance between the lines l1 and l2 given by \[\overrightarrow{r} = \hat{i} + 2 \hat{j} - 4 \hat{k} + \lambda\left( 2 \hat{i} + 3 \hat{j} + 6 \hat{k} \right) \text{ and } , \overrightarrow{r} = 3 \hat{i} + 3 \hat{j} - 5 \hat{k} + \mu\left( 2 \hat{i} + 3 \hat{j} + 6 \hat{k} \right)\]
The perpendicular distance of the point P (1, 2, 3) from the line \[\frac{x - 6}{3} = \frac{y - 7}{2} = \frac{z - 7}{- 2}\] is
If a line makes angle \[\frac{\pi}{3} \text{ and } \frac{\pi}{4}\] with x-axis and y-axis respectively, then the angle made by the line with z-axis is
If the lines represented by kx2 − 3xy + 6y2 = 0 are perpendicular to each other, then
Find the position vector of a point A in space such that `vec"OA"` is inclined at 60º to OX and at 45° to OY and `|vec"OA"|` = 10 units.
In the vector equation \[\vec{r}=\vec{a}+\lambda\vec{b}\], what does \[\vec{r}\] represent?
For the point \[(5,2,-4)\], what is its position vector \[\vec{a}\]?
Which Cartesian equation is obtained for the line through \[(5,2,-4)\] parallel to \[3\hat{i}+2\hat{j}-8\hat{k}\]?
