Foot of the Perpendicular from a Point on a Plane

Q.

Perpendicular are drawn from points on the line $\frac{x+2}{2}=\frac{y+1}{-1}=\frac{z}{3}$ to the plane x + y + z = 3. The feet of perpendiculars lie on the line

1. $\frac{x}{5}=\frac{y-1}{8}=\frac{z-2}{-13}$

2. $\frac{x}{2}=\frac{y-1}{3}=\frac{z-2}{-5}$

3. $\frac{x}{4}=\frac{y-1}{3}=\frac{z-2}{-7}$

4. $\frac{x}{2}=\frac{y-1}{-7}=\frac{z-2}{5}$

Q.

The equation of the line bisecting perpendicularly the segment joining the points $(-4,6)$ and $(8,8)$ is:

1. $6x+y-19=0$

2. $y=7$

3. $6x+2y-19=0$

4. $x+2y-7=0$

Q. The length and foot of the perpendicular from the point (7, 14, 5) to the plane 2x + 4y - z = 2, are

1. $\sqrt{21},(1,2,8)$
2. $3\sqrt{21},(3,2,8)$
3. $21\sqrt{3},(1,2,8)$
4. $3\sqrt{21},(1,2,8)$

Q.

The coordinates of the foot of the perpendicular drawn from the point

P(3, 4, 5) on the yz-plane are

1. (3, 4, 0)
2. (0, 4, 5)
3. (3, 0, 5)
4. (3, 0, 0)

Q. The maximum value of Z = 4x + 2y subjected to the constraints 2x + 3y ≤ 18, x + y ≥ 10; x, y ≥ 0 is
(a) 36
(b) 40
(c) 20
(d) none of these

Q.

A plane passes through a fixed point (a, b, c). Then the locus of the foot of the perpendicular to it from the origin is

1. $x^2+y^2+z^2-2ax-2by-2cz=0$

2. $2x^2+2y^2+2z^2-ax-by-cz=0$

3. $x^2+y^2+z^2-4ax-4by-4cz=0$

4. $x^2+y^2+z^2-ax-by-cz=0$

Q. Let $P(a,b,c)$ be any point on the plane $3x+2y+z=7.$ Then the least value of $2(a^2+b^2+c^2)$ is

Q.

The foot of the perpendicular drawn from the (- 1, - 3, - 5) to a plane is (2, 4, 6). The equation of the plane is:

1. 3x + 7y + 11z = 44

2. x + y + z = 44

3. 3x + 7y + 11z = 100

4. 3x + 7y + 11z = 88

Q.

Find the equation of the parabola that satisfies the following conditions: Vertex (0, 0) focus (–2, 0)

Q. Find the coordinates of the foot of the perpendicular drawn from the origin to the plane 2x – 3y + 4z – 6 = 0.

1. $\left(12,-18,24\right)$
2. $\left(\frac{12}{29},-\frac{18}{29},\frac{24}{29}\right)$
3. $\left(\frac{12}{\sqrt{29}},-\frac{18}{\sqrt{29}},\frac{24}{\sqrt{29}}\right)$
4. $(29,-29,29)$

Q. If the line $\frac{x+1}{-1}=\frac{y-1}{2}=\frac{z-2}{1}$ lies on the plane $nx+my-2z=4,$ then $m-n=$

Q.

If the distance of the point P (1, -2, 1) from the plane $x+2y-2z=\alpha$ where $\alpha >0$, is 5, then the foot of the perpendicular form P to the plane is

1. $(\frac{8}{3},\frac{4}{3},-\frac{7}{3})$

2. $(\frac{4}{3},\frac{4}{3},\frac{1}{3})$

3. $(\frac{1}{3},\frac{2}{3},\frac{10}{3})$

4. $(\frac{2}{3},\frac{1}{3},\frac{5}{2})$

Q. 7. The equation of plane containing the lines 2x-5y+z=3, x+y+4z=5 and parallel to the plane x+3y+6z=1 is

Q. The co-ordinates of the foot of the perpendicular drawn from the origin to a plane is (2, 4, -3). The equation of the plane is

1. 2x-4y-3z=29
2. 2x-4y+3z=29
3. 2x+4y-3z=29
4. None of these

Q.

The position vector of mid-point of joining the points (2, – 1, 3) and (4, 3, –5) is:

1. 

2. 

3. 

4. 

Q. Two vertices of the parallelogram ABCD are A(8, 14, 12) and B(4, 6, 4) and its diagonals intersect at (3, 5, 5).Find the coordinates of vertices C and D.

Q. Find the distance of the point (2, 4, −1) from the line $\frac{x+5}{1}=\frac{y+3}{4}=\frac{z-6}{-9}$. [NCERT EXEMPLAR]

Q.

What will be the perpendicular distance of a point P (7, 5 , -2) from the plane -2x +7y - 4z + 5 = 0 ?

1. $\frac{23}{\sqrt{69}}$

2. $\frac{34}{\sqrt{69}}$

3. $\frac{57}{\sqrt{69}}$

4. $\frac{62}{\sqrt{69}}$

Q. Find the vector and Cartesian forms of the equation of the plane passing through the point (1, 2, −4) and parallel to the lines

$\overrightarrow{r}=\left(\hat{i}+2\hat{j}-4\hat{k}\right)+\lambda \left(2\hat{i}+3\hat{j}+6\hat{k}\right)$ and $\overrightarrow{r}=\left(\hat{i}-3\hat{j}+5\hat{k}\right)+\mu \left(\hat{i}+\hat{j}-\hat{k}\right)$. Also, find the distance of the point (9, −8, −10) from the plane thus obtained. [CBSE 2014]

Q.

Find the vector and the Cartesian equations of the line that passes through the points (3, −2, −5), (3, −2, 6).

Q.

Find the equation of a line parallel to x-axis and passing through the origin.

Q. The co-ordinates of the foot of the perpendicular drawn from the origin to a plane is (2, 4, -3). The equation of the plane is

1. 2x-4y-3z=29
2. 2x-4y+3z=29
3. 2x+4y-3z=29
4. None of these

Q. Find the distance of the point (1, −2, 3) from the plane x − y + z = 5 measured along a line parallel to $\frac{x}{2}=\frac{y}{3}=\frac{z}{-6}.$

Q.

Find the equation of the parabola that satisfies the following conditions: Vertex (0, 0); focus (3, 0)

Q. find the equation of the plane through the line of intersection of the plane x+y+z=1 and 2x+3y+4z=5 which is perpendicular to the plane x-y+z=0.
then find the distance of the plane thus obtained from the point A(1, 3, 6).







=

Q. The maximum value of Z = 4x + 3y subjected to the constraints 3x + 2y ≥ 160, 5x + 2y ≥ 200, x + 2y ≥ 80; x, y ≥ 0 is
(a) 320
(b) 300
(c) 230
(d) none of these

Q. Find the vector equation of a line passing through the point with position vector $\hat{i}-2\hat{j}-3\hat{k}$and parallel to the line joining the points with position vectors $\hat{i}-\hat{j}+4\hat{k}\mathrm{and}2\hat{i}+\hat{j}+2\hat{k}.$ Also, find the cartesian equivalent of this equation.

Q. The locus of foot of the perpendiculars drawn from the line $\frac{x-2}{2}=\frac{y-1}{1}=\frac{z}{3}$ to the plane $x+y+z=1$ is

1. $\frac{x-1}{0}=\frac{y}{-1}=\frac{z+1}{1}$
2. $\frac{x-\frac{4}{3}}{0}=\frac{y-\frac{1}{3}}{1}=\frac{z+\frac{2}{3}}{-1}$
3. $\frac{x}{0}=\frac{y-1}{-1}=\frac{z+1}{1}$
4. $\frac{x-\frac{4}{3}}{0}=\frac{y-\frac{1}{3}}{-1}=\frac{z+\frac{2}{3}}{1}$

Q. Write the equation of the plane parallel to XOY- plane and passing through the point (2, −3, 5).

Q. Maximize Z = −x₁ + 2x₂
Subject to
$$\begin{gathered} -x_1+3x_2\le 10 \\ {x}_1+x_2\le 6 \\ {x}_1-x_2\le 2 \\ {x}_1,x_2\ge 0 \end{gathered}$$