Advertisements
Advertisements
Question
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)`
Solution
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
RELATED QUESTIONS
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.
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 vector equation of a plane which is at a distance of 5 units from the origin and its normal vector is `2hati-3hatj+6hatk`
Find the vector equation of a line passing through the points A(3, 4, –7) and B(6, –1, 1).
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.
3y + 4z – 6 = 0
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 equations of the plane passing throuh the points (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 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).
Vector equation of a line which passes through a point (3, 4, 5) and parallels to the vector `2hati + 2hatj - 3hatk`.
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 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")`.
