Advertisements
Advertisements
प्रश्न
Find the equation of the line passing through the point (1, −1, 1) and perpendicular to the lines joining the points (4, 3, 2), (1, −1, 0) and (1, 2, −1), (2, 1, 1).
उत्तर
The direction ratios of the line joining the points (4, 3, 2), (1, -1, 0) and (1, 2, -1), (2, 1, 1) are -3,-4,-2 and 1, -1,2 ,respectively.
Let:
\[\overrightarrow{b_1} = - 3 \hat{i} - 4 \hat{j} - 2 \hat{k} \]
\[ \overrightarrow{b_2} = \hat{i} - \hat{j} + 2 \hat{k}\]
Since the required line is perpendicular to the lines parallel to the vectors
\[\overrightarrow{b_1} = - 3 \hat{i} - 4 \hat{j} - 2 \hat{k} \text{ and } \overrightarrow{b_2} = \hat{i} - \hat{j} + 2 \hat{k} \] it is parallel to the vector
\[\overrightarrow{b} = \overrightarrow{b_1} \times \overrightarrow{b_2}\]
Now,
\[\overrightarrow{b} = \overrightarrow{b_1} \times \overrightarrow{b_2} \]
\[ = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ - 3 & - 4 & - 2 \\ 1 & - 1 & 2\end{vmatrix}\]
\[ = - 10 \hat{i} + 4 \hat{j} + 7 \hat{k}\]
So, the direction ratios of the required line are proportional to
-10, 4, 7.
The equation of the required line passing through the point (1,-1, 1) and having direction ratios proportional to -10, 4, 7 is
\[\frac{x - 1}{- 10} = \frac{y + 1}{4} = \frac{z - 1}{7}\]
APPEARS IN
संबंधित प्रश्न
If a line drawn from the point A( 1, 2, 1) is perpendicular to the line joining P(1, 4, 6) and Q(5, 4, 4) then find the co-ordinates of the foot of the perpendicular.
Show that the line joining the origin to the point (2, 1, 1) is perpendicular to the line determined by the points (3, 5, – 1), (4, 3, – 1).
ABCD is a parallelogram. The position vectors of the points A, B and C are respectively, \[4 \hat{ i} + 5 \hat{j} -10 \hat{k} , 2 \hat{i} - 3 \hat{j} + 4 \hat{k} \text{ and } - \hat{i} + 2 \hat{j} + \hat{k} .\] Find the vector equation of the line BD. Also, reduce it to cartesian form.
Find the cartesian equation of a line passing through (1, −1, 2) and parallel to the line whose equations are \[\frac{x - 3}{1} = \frac{y - 1}{2} = \frac{z + 1}{- 2}\] Also, reduce the equation obtained in vector form.
Find the points on the line \[\frac{x + 2}{3} = \frac{y + 1}{2} = \frac{z - 3}{2}\] at a distance of 5 units from the point P (1, 3, 3).
Show that the points whose position vectors are \[- 2 \hat{i} + 3 \hat{j} , \hat{i} + 2 \hat{j} + 3 \hat{k} \text{ and } 7 \text{ i} - \text{ k} \] are collinear.
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).
Show that the line through the points (4, 7, 8) and (2, 3, 4) is parallel to the line through the points (−1, −2, 1) and, (1, 2, 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 \[\frac{x - 5}{7} = \frac{y + 2}{- 5} = \frac{z}{1} \text{ and } \frac{x}{1} = \frac{y}{2} = \frac{z}{3}\] are perpendicular to each other.
Find the vector equation of the line passing through the point (2, −1, −1) which is parallel to the line 6x − 2 = 3y + 1 = 2z − 2.
Show that the lines \[\frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7} \text{ and } \frac{x - 2}{1} = \frac{y - 4}{3} = \frac{z - 6}{5}\] intersect. 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 (1, 0, 0) from the line \[\frac{x - 1}{2} = \frac{y + 1}{- 3} = \frac{z + 10}{8}.\] Also, find the coordinates of the foot of the perpendicular and the equation of the perpendicular.
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 (1, 2, −3) to the line \[\frac{x + 1}{2} = \frac{y - 3}{- 2} = \frac{z}{- 1} .\]
Find the distance of the point (2, 4, −1) from the line \[\frac{x + 5}{1} = \frac{y + 3}{4} = \frac{z - 6}{- 9}\]
Find the shortest distance between the following pairs of lines whose vector equations are: \[\overrightarrow{r} = \left( 3 \hat{i} + 5 \hat{j} + 7 \hat{k} \right) + \lambda\left( \hat{i} - 2 \hat{j} + 7 \hat{k} \right) \text{ and } \overrightarrow{r} = - \hat{i} - \hat{j} - \hat{k} + \mu\left( 7 \hat{i} - 6 \hat{j} + \hat{k} \right)\]
Find the shortest distance between the following pairs of lines whose vector equations are: \[\overrightarrow{r} = \left( 1 - t \right) \hat{i} + \left( t - 2 \right) \hat{j} + \left( 3 - t \right) \hat{k} \text{ and } \overrightarrow{r} = \left( s + 1 \right) \hat{i} + \left( 2s - 1 \right) \hat{j} - \left( 2s + 1 \right) \hat{k} \]
Find the shortest distance between the following pairs of lines whose vector equations are: \[\overrightarrow{r} = \left( \lambda - 1 \right) \hat{i} + \left( \lambda + 1 \right) \hat{j} - \left( 1 + \lambda \right) \hat{k} \text{ and } \overrightarrow{r} = \left( 1 - \mu \right) \hat{i} + \left( 2\mu - 1 \right) \hat{j} + \left( \mu + 2 \right) \hat{k} \]
Find the shortest distance between the following pairs of lines whose cartesian equations are: \[\frac{x - 3}{1} = \frac{y - 5}{- 2} = \frac{z - 7}{1} \text{ and } \frac{x + 1}{7} = \frac{y + 1}{- 6} = \frac{z + 1}{1}\]
Find the shortest distance between the following pairs of parallel lines whose equations are: \[\overrightarrow{r} = \left( \hat{i} + 2 \hat{j} + 3 \hat{k} \right) + \lambda\left( \hat{i} - \hat{j} + \hat{k} \right) \text{ and } \overrightarrow{r} = \left( 2 \hat{i} - \hat{j} - \hat{k} \right) + \mu\left( - \hat{i} + \hat{j} - \hat{k} \right)\]
Find the shortest distance between the following pairs of parallel lines whose equations are: \[\overrightarrow{r} = \left( \hat{i} + \hat{j} \right) + \lambda\left( 2 \hat{i} - \hat{j} + \hat{k} \right) \text{ and } \overrightarrow{r} = \left( 2 \hat{i} + \hat{j} - \hat{k} \right) + \mu\left( 4 \hat{i} - 2 \hat{j} + 2 \hat{k} \right)\]
Find the shortest distance between the lines \[\overrightarrow{r} = \left( \hat{i} + 2 \hat{j} + \hat{k} \right) + \lambda\left( \hat{i} - \hat{j} + \hat{k} \right) \text{ and } , \overrightarrow{r} = 2 \hat{i} - \hat{j} - \hat{k} + \mu\left( 2 \hat{i} + \hat{j} + 2 \hat{k} \right)\]
Find the shortest distance between the lines \[\overrightarrow{r} = \hat{i} + 2 \hat{j} + 3 \hat{k} + \lambda\left( \hat{i} - 3 \hat{j} + 2 \hat{k} \right) \text{ and } \overrightarrow{r} = 4 \hat{i} + 5 \hat{j} + 6 \hat{k} + \mu\left( 2 \hat{i} + 3 \hat{j} + \hat{k} \right)\]
The lines `x/1 = y/2 = z/3 and (x - 1)/-2 = (y - 2)/-4 = (z - 3)/-6` are
If a line makes angles α, β and γ with the axes respectively, then cos 2 α + cos 2 β + cos 2 γ =
The projections of a line segment on X, Y and Z axes are 12, 4 and 3 respectively. The length and direction cosines of the line segment are
If the lines represented by kx2 − 3xy + 6y2 = 0 are perpendicular to each other, then
P is a point on the line joining the points A(0, 5, −2) and B(3, −1, 2). If the x-coordinate of P is 6, then its z-coordinate is ______.
Find the vector equation of the lines passing through the point having position vector `(-hati - hatj + 2hatk)` and parallel to the line `vecr = (hati + 2hatj + 3hatk) + λ(3hati + 2hatj + hatk)`.
Equation of a line passing through (1, 1, 1) and parallel to z-axis is ______.
A line passes through the point (2, – 1, 3) and is perpendicular to the lines `vecr = (hati + hatj - hatk) + λ(2hati - 2hatj + hatk)` and `vecr = (2hati - hatj - 3hatk) + μ(hati + 2hatj + 2hatk)` obtain its equation.
