Advertisements
Advertisements
प्रश्न
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)`
उत्तर
Let the position vectors of the three points be `veca= hati+hatj-2hatk , vecb=2hati-hatj+hatk and vecc=hati+2hatj+hatk`
So, the equation of the plane passing through the points `vec a,vecb and vecc` is
`(vecr-veca).[(vecb-vecc)xx(vecc-veca)]=0`
`[vecr-(hati+hatj-2hatk)][(hati-3hatj)xx(hatj+3hatk)]=0`
`[vecr-(hati+hatj-2hatk)](hatk-3hatj-9hati)=0`
`vecr(-9hati-3hatj+hatk)-(-9-3-2)=0`
`vecr(-9hati-3hatj+hatk)+14=0`
`vecr(9hati+3hatj-hatk)=14 ............(1)`
So, the vector equation of the plane is `vecr(9hati+3hatj-hatk)=14`
The equation of the given line is `vecr=3hati-hatj-hatk lambda +(2hati-2hatj+hatk)`
Position vector of any point on the give line is
`vecr=(3+2lambda)hati+(-1-2lambda)hatj+(-1+lambda)hatk ...................(2)`
the point(2) lies on plane (1) if ,
`[(3+2lambda)hati+(-1-2lambda)hatj+(-1+lambda)hatk](9hati+3hatj-hatk)=14`
`9(3+2lambda)+3(-1-2lambda)-(-1+lambda)=14`
`11 lambda+25=14`
`lambda=-1`
Putting `lambda=-1` in (2) we have
`vecr=(3-2)hati+(-1+2)hatj+(-1-1)hatk`
`=hati+hatj-2hatk`
Thus, the point of intersection of the given line and the plane (1) is `hati+hatj-2hatk`
APPEARS IN
संबंधित प्रश्न
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.
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.
Write the vector equation of the plane, passing through the point (a, b, c) and parallel to the plane `vec r.(hati+hatj+hatk)=2`
Find the vector equation of the plane with intercepts 3, –4 and 2 on x, y and z-axis respectively.
Find the equation of the plane which contains the line of intersection of the planes
`vecr.(hati-2hatj+3hatk)-4=0" and"`
`vecr.(-2hati+hatj+hatk)+5=0`
and whose intercept on x-axis is equal to that of on y-axis.
The x-coordinate of a point of the line joining the points P(2,2,1) and Q(5,1,-2) is 4. Find its z-coordinate
Find the Cartesian equation of the following planes:
`vecr.(hati + hatj-hatk) = 2`
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 vector and Cartesian equation of the planes that passes through the point (1, 0, −2) and the normal to the plane is `hati + hatj - hatk`
Find the cartesian form of the equation of the plane `bar r=(hati+hatj)+s(hati-hatj+2hatk)+t(hati+2hatj+hatj)`
Find the equation of the plane through the line of intersection of `vecr*(2hati-3hatj + 4hatk) = 1`and `vecr*(veci - hatj) + 4 =0`and perpendicular to the plane `vecr*(2hati - hatj + hatk) + 8 = 0`. Hence find whether the plane thus obtained contains the line x − 1 = 2y − 4 = 3z − 12.
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 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 vector equation of the line passing through (1, 2, 3) and parallel to the planes \[\vec{r} \cdot \left( \hat{i} - \hat{j} + 2 \hat{k} \right) = 5 \text{ and } \vec{r} \cdot \left( 3 \hat{i} + \hat{j} + \hat{k} \right) = 6 .\]
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 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).
The vector equation of the line `(x - 5)/3 = (y + 4)/7 = (z - 6)/2` is ______.
The vector equation of the line through the points (3, 4, –7) and (1, –1, 6) 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")`.
Find the vector and the cartesian equations of the plane containing the point `hati + 2hatj - hatk` and parallel to the lines `vecr = (hati + 2hatj + 2hatk) + s(2hati - 3hatj + 2hatk)` and `vecr = (3hati + hatj - 2hatk) + t(hati - 3hatj + hatk)`
