Advertisements
Advertisements
प्रश्न
Find the vector equation of the plane that contains the lines `vecr = (hat"i" + hat"j") + λ (hat"i" + 2hat"j" - hat"k")` and the point (–1, 3, –4). Also, find the length of the perpendicular drawn from the point (2, 1, 4) to the plane thus obtained.
Advertisements
उत्तर
Let the vector equation of the required plane be `vec"r" . vec"n" = d`
The plane contains the line `vec"r" = hat"i" + hat"j" + λ (hat"i" + 2hat"j" - hat"k")`
Since the plane passes through point A and B. So `vec"n"` will be parallel to vector `vec"AB" xx (hat"i" + 2hat"j" - hat"k")`
`vec"AB" = vec"OB" -vec"OA"`
= `(-hat"i" + 3hat"j" - 4hat"k") (hat"i" + hat"j")`
= `-2hat"i" + 2hat"j" - 4hat"k"`
`vec"AB" xx (hat"i" + 2hat"j" - hat"k") = |(hat"i",hat"j",hat"k"),(1,2,-1),(-2,2, -4)|`
= `hat"i" (-8 + 2) - hat"j" (-4-2) + hat"k" (2+4)`
= `-6hat"i" + 6hat"j" + 6hat"k"`
which is a normal vector to the plane.
So the equation of plane will be `vec"r" . (-6hat"i" + 6hat"j" + 6hat"k") = d`
∴ it passes through (1, 1, 0) so `(hat"i" + hat"j"). (-6hat"i" + 6hat"j" + 6hat"k") = d or, d = 0`
equation of plane is `vecr . (-6hat"i" + 6hat"j" + 6hat"k") = 0`
`vecr (hat"i" - hat"j" -hat"k") = 0`
in Cartesian plane,
`(xhat"i" + yhat"j" + zhat"k") . (hat"i" - hat"j" -hat"k") = 0`
x -y - z = 0
So, the perpendicular distance of the plane from the point (2, 1, 4) is ` = |(2 -1-4)/sqrt(1^2 + (-1)^2 + (-1)^2)| = |(-3)/sqrt(3)| = sqrt(3) "unit"`.
संबंधित प्रश्न
Find the vector equation of the plane passing through a point having position vector `3 hat i- 2 hat j + hat k` and perpendicular to the vector `4 hat i + 3 hat j + 2 hat k`
Find the vector equation of the plane passing through the points `hati +hatj-2hatk, hati+2hatj+hatk,2hati-hatj+hatk`. Hence find the cartesian equation of the plane.
Find the vector equation of the plane which is at a distance of 5 units from the origin and which is normal to the vector `2hati + hatj + 2hatk.`
Parametric form of the equation of the plane is `bar r=(2hati+hatk)+lambdahati+mu(hat i+2hatj+hatk)` λ and μ are parameters. Find normal to the plane and hence equation of the plane in normal form. Write its Cartesian form.
Find the vector equation of the plane passing through three points with position vectors ` hati+hatj-2hatk , 2hati-hatj+hatk and hati+2hatj+hatk` . Also find the coordinates of the point of intersection of this plane and the line `vecr=3hati-hatj-hatk lambda +(2hati-2hatj+hatk)`
Find the vector equation of the plane with intercepts 3, –4 and 2 on x, y and z-axis respectively.
Find the Cartesian equation of the following planes:
`vecr.(2hati + 3hatj-4hatk) = 1`
Find the Cartesian equation of the following planes:
`vecr.[(s-2t)hati + (3 - t)hatj + (2s + t)hatk] = 15`
In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.
2x + 3y + 4z – 12 = 0
In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.
5y + 8 = 0
Find the cartesian form of the equation of the plane `bar r=(hati+hatj)+s(hati-hatj+2hatk)+t(hati+2hatj+hatj)`
The Cartesian equation of the line is 2x - 3 = 3y + 1 = 5 - 6z. Find the vector equation of a line passing through (7, –5, 0) and parallel to the given line.
Find the image of a point having the position vector: `3hati - 2hatj + hat k` in the plane `vec r.(3hati - hat j + 4hatk) = 2`
Find the vector and Cartesian equations of the line passing through (1, 2, 3) and parallel to the planes \[\vec{r} \cdot \left( \hat{i} - \hat{j} + 2 \hat{k} \right) = 5 \text{ and } \vec{r} \cdot \left( 3 \hat{i} + \hat{j} + 2 \hat{k} \right) = 6\]
Find the vector and Cartesian forms of the equation of the plane passing through the point (1, 2, −4) and parallel to the lines \[\vec{r} = \left( \hat{i} + 2 \hat{j} - 4 \hat{k} \right) + \lambda\left( 2 \hat{i} + 3 \hat{j} + 6 \hat{k} \right)\] and \[\vec{r} = \left( \hat{i} - 3 \hat{j} + 5 \hat{k} \right) + \mu\left( \hat{i} + \hat{j} - \hat{k} \right)\] Also, find the distance of the point (9, −8, −10) from the plane thus obtained.
Find the equation of the plane passing through the intersection of the planes `vec(r) .(hat(i) + hat(j) + hat(k)) = 1"and" vec(r) . (2 hat(i) + 3hat(j) - hat(k)) +4 = 0 `and parallel to x-axis. Hence, find the distance of the plane from x-axis.
Find the Cartesian equation of the plane, passing through the line of intersection of the planes `vecr. (2hati + 3hatj - 4hatk) + 5 = 0`and `vecr. (hati - 5hatj + 7hatk) + 2 = 0` intersecting the y-axis at (0, 3).
Find the vector and cartesian equation of the plane passing through the point (2, 5, - 3), (-2, -3, 5) and (5, 3, -3). Also, find the point of intersection of this plane with the line passing through points (3, 1, 5) and (-1, -3, -1).
Find the vector and Cartesian equations of the plane passing through the points (2, 2 –1), (3, 4, 2) and (7, 0, 6). Also find the vector equation of a plane passing through (4, 3, 1) and parallel to the plane obtained above.
The Cartesian equation of the plane `vec"r" * (hat"i" + hat"j" - hat"k")` = 2 is ______.
The vector equation of the line `(x - 5)/3 = (y + 4)/7 = (z - 6)/2` is `vec"r" = 5hat"i" - 4hat"j" + 6hat"k" + lambda(3hat"i" + 7hat"j" + 2hat"k")`.
