# Equation of a Plane Passing through Three Points

## Trending Questions

**Q.**

A plane passes through the points $A(1,2,3),B(2,3,1)$ and $C(2,4,2)$. If $O$ is the origin and $P$ is $(2,\u20131,1)$, then the projection of vector$\left(OP\right)$ on this plane is of length:

$\sqrt{\frac{2}{5}}$

$\sqrt{\frac{2}{3}}$

$\sqrt{\frac{2}{11}}$

$\sqrt{\frac{2}{7}}$

**Q.**

If a plane cuts off intercepts -6, 3, 4 from the co-ordinate axes, then the length of the perpendicular from the origin to the plane is

**Q.**The direction ratio of the normal to the plane passing through the points (1, 2, −3), (−1, −2, 1) and parallel to x−22=y+13=z4 is

- (2, 3, 4)
- (14, −8, −1)
- (−2, 0, −3)
- (1, −2, −3)

**Q.**

Let P be the image of the point (3, 1, 7) with respect to the plane x - y + z = 3. Then, the equation of the plane passing through P and containing the straight line x1=y2=z1 is

x + y - 3z = 0

3x + z = 0

x - 4y + 7z = 0

2x - y = 0

**Q.**

Let $P$ be a plane passing through the points $(2,1,0),(4,1,1)$and $(5,0,1)$ and $R$ be any point$(2,1,6).$ Then the image of $R$ in the plane $P$ is:

$(6,5,2)$

$(6,5,-2)$

$(4,3,2)$

$(3,4,-2)$

**Q.**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.

**Q.**

The distance of the point $\left(3,8,2\right)$ from the line $\frac{x-1}{2}=\frac{y-3}{4}=\frac{z-2}{3}$ measured parallel to the plane $3x+2y-2z=0$ is

$2$

$3$

$6$

$7$

**Q.**

Find the equation of plane passing through the points P (1, 1, 1) , Q (3, -1, 2) and R (-3, 5 , -4).

6x + 16y +16z - 44 = 0

x - 2y +16z - 13 = 0

6x + 6y +16z - 28 = 0

x + 6y +6z - 22 = 0

**Q.**A plane meets the co-ordinates axes in A, B, C and (α, β, γ) is the centroid of the ΔABC. Then the equation of the plane is :

- 3xα+3yβ+3zγ=1
- αx+βy+γz=1
- xα+yβ+zγ=1
- xα+yβ+zγ=3

**Q.**If →a+→b+→c=α→d, →b+→c+→d=β→a and →a, →b, →c are non-coplanar, then the sum of →a+→b+→c+→d=

- →0
- (β−1)⋅→d+(α−1)⋅→a
- (α−1)⋅→d−(β−1)⋅→a
- (α−1)⋅→d+(β−1)⋅→a

**Q.**

If ${\mathrm{z}}_{1}=1+\mathrm{i},{\mathrm{z}}_{2}=\u20132+3\mathrm{i}$ and ${\mathrm{z}}_{3}=\frac{{\mathrm{a}}_{\mathrm{i}}}{3}$ where ${\mathrm{i}}^{2}=\u20131$, are collinear then the value of $\mathrm{a}$ is

$-1$

$3$

$4$

$5$

**Q.**

Find the vector and Cartesian equations of the planes.

that passes through the point (1, 0, -2) and the normal to the plane is ^i+^j−^k.

that passes through the point (1, 4, 6) and the normal to the plane is ^i−2^j+^k.

**Q.**A variable plane passes through a fixed point (3, 2, 1) and meets x, y and z axes at A, B and C respectively. A plane is drawn parallel to yz−plane through A, a second plane is drawn parallel zx−plane through B and a third plane is drawn parallel to xy−plane through C. Then the locus of the point of intersection of these three planes, is :

- x+y+z=6
- x3+y2+z1=1
- 3x+2y+1z=1
- 1x+1y+1z=116

**Q.**

If four points P (1, 2, 3) , Q (2, 3, 4), R(3, 4, 5), S(4, 5, k) are coplanar then the value of k is / are -

6

4

5

Any real number.

**Q.**A plane passes through the points A(1, 2, 3), B(2, 3, 1) and C(2, 4, 2). If O is the origin and P is (2, −1, 1), then the projection of −−→OP on this plane is of length :

- √23
- √25
- √211
- √27

**Q.**

Find the vector and Cartesian equations of the plane,

that passes through the point (1, 4, 6) and the normal to the plane is ^i−2^j+^k.

**Q.**Let P be the image of the point (3, 1, 7) with respect to the plane x−y+z=3. Then the equation of the plane passing through P and containing the straight line x1=y2=z1 is

- x+y−3z=0
- x−4y+7z=0
- 2x−y=0
- 3x+z=0

**Q.**A perpendicular is drawn from the point A(1, 8, 4) to the line joining the points B(0, −1, 3) and C(2, −3, −1). The co-ordinates of the foot of the perpendicular is:

- (53, 23, 193)
- (−53, 23, 193)
- (53, −23, 193)
- (53, 23, −193)

**Q.**Find the equation of the plane passing through the points (1, 2, 3), (3, 5, 7) and (4, 3, 1)

**Q.**Show that the line through points (4, 7, 8) and (2, 3, 4) is parallel to the line through the points (−1, −2, 1) and (1, 2, 5).

**Q.**A variable plane intersects the coordinate axes at A, B, C and is at a constant distance p from O(0, 0, 0). Then the locus of the centroid of the tetrahedron OABC is

- 1x2+1y2+1z2=16p2
- 1x2+1y2+1z2=16p2
- 1x2+1y2+1z2=1p2
- 1x2+1y2+1z2=4p2

**Q.**

The position vectors of $P$ and $Q$ are respectively, $a$ and $b$. If $R$ is point on $PQ$ such that $PR=5PQ$, then the position vector $R$ is?

$5b\u20134a$

$5b+4a$

$4a-5b$

$4a+5b$

**Q.**Equation of the plane through three points A, B, C with position vectors −6→i+3→j+2→k, 3→i−2→j+4→k, 5→i+7→j+3→k is

- →r⋅(→i−→j−7→k)+23=0
- →r⋅(→i+→j+7→k)=23
- →r⋅(→i+→j−7→k)+23=0
- →r⋅(→i−→j−7→k)=23

**Q.**A plane passes through the points A(1, 2, 3), B(2, 3, 1) and C(2, 4, 2). If O is the origin and P is (2, −1, 1), then the projection of −−→OP on this plane is of length :

- √25
- √23
- √211
- √27

**Q.**A plane moves such that its distance from the origin is a constant p. If it intersects the coordinate axes at A, B,

C then the locus of the centroid of the triangle ABC is

- 1x2+1y2+1z2=1p2
- 1x2+1y2+1z2=9p2
- 1x2+1y2+1z2=2p2
- 1x2+1y2+1z2=4p2

**Q.**

If four points P (1, 2, 3) , Q (2, 3, 4), R(3, 4, 5), S(4, 5, k) are coplanar then the value of k is / are -

6

4

5

Any real number.

**Q.**Consider three vectors →p=i+j+k, →q=2i+4j−k and →r=i+j+3k. If p, q and r denotes the position vector of three non-collinear points, then the equation of the plane containing these points is

- 2x−3y+1=0
- x−3y+2z=0
- 3x−y+z−3=0
- 3x−y−2=0

**Q.**The perpendicular distance from the point (3, 1, 1) on the plane passing through the point (1, 2, 3) and containing the line, →r=^i+^j+λ(2^i+^j+4^k), is:

- 1√11
- 4√41
- 0
- 3√11

**Q.**Vector equation of the plane →r=^i−^j+λ(^i−^j+^k)+μ(^i−2^j+3^k)

- −^i−2^j−^k
- →r.(5^i+2^j−3^k)=7
- →r.(5^i−2^j−3^k)=7
- →r.(5^i+2^j+3^k)=7

**Q.**

Find the Cartesian equation of the following plane:

r.(2^i+3^j−4^k)=1