Free Three Dimensional Geometry 02 Practice Test - 12th Grade - Commerce

Question 1

If the perpendicular distance of a point other than the origin from the plane x+y+z=p is equal to the distance of the plane from the origin, then the coordinates of the point are

A. (p, 2 p, 0)

B. (0, 2p, -p)

C. (2p, p, -p)

D. (2p, -p, 2p)

SOLUTION

Solution : C

The perpendicular distance of the origin (0, 0, 0) from the plane x + y + z = p is $∣ ∣ ∣ \frac{- p}{\sqrt{1 + 1 + 1}} ∣ ∣ ∣ = \frac{| p |}{\sqrt{3}}$
If the coordinates of P are (x, y, z), then we must have
$∣ ∣ ∣ \frac{x + y + z - p}{\sqrt{3}} ∣ ∣ ∣ = \frac{| p |}{\sqrt{3}} \Rightarrow | x + y + z - p | = | p |$
Which is satisfied by (c)

Question 2

The shortest distance from the plane 12x + 4y + 3z = 327 to the sphere $x^{2} + y^{2} + z^{2} + 4 x - 2 y - 6 z = 155$ is

11 \frac{3}{4}

13

39

26

SOLUTION

Solution : B

The centre of the sphere is (-2, 1, 3) and its radius is $\sqrt{4 + 1 + 9 + 155} = 13$ .
Length of the perpendicular from the centre of the sphere on the plane is $∣ ∣ ∣ \frac{- 24 + 4 + 9 - 327}{\sqrt{144 + 16 + 9}} ∣ ∣ ∣ = \frac{338}{13} = 26$
So the plane is outside the sphere and the required distance is equal to 26 - 13 = 13.

Question 3

The ratio in which yz- plane divides the segment joining the points (-2, 4, 7) and (3, -5, 8) is

A. 2 : 3

B. 3 : 2

C. 4 : 5

D. -7 : 8

SOLUTION

Solution : A

let YZ-plane divide the segment joining (-2, 4, 7) and (3, -5, 8) in the ratio $λ : 1$ . Then $\Rightarrow \frac{3 λ - 2}{λ + 1} = 0 \Rightarrow λ = \frac{2}{3}$ and the required ratio is 2 : 3

Question 4

If $(l_{1}, m_{1}, n_{1})$ and $(l_{2}, m_{2}, n_{2},)$ are d.c.'s of $¯ ¯¯¯¯¯¯¯¯ ¯ O A,$ $¯ ¯¯¯¯¯¯ ¯ O B$ such that $∠ A O B = θ$ where ‘O’ is the origin, then the d.c.’s of the internal bisector of the angle $∠ A O B$ are

\frac{l_{1} + l_{2}}{2 s i n \frac{θ}{2}}, \frac{m_{1} + m_{2}}{2 s i n \frac{θ}{2}}, \frac{n_{1} + n_{2}}{2 s i n \frac{θ}{2}}

\frac{l_{1} + l_{2}}{2 c o s \frac{θ}{2}}, \frac{m_{1} + m_{2}}{2 c o s \frac{θ}{2}}, \frac{n_{1} + n_{2}}{2 c o s \frac{θ}{2}}

\frac{l_{1} - l_{2}}{2 s i n \frac{θ}{2}}, \frac{m_{1} - m_{2}}{2 s i n \frac{θ}{2}}, \frac{n_{1} - n_{2}}{2 s i n \frac{θ}{2}}

\frac{l_{1} - l_{2}}{2 c o s \frac{θ}{2}}, \frac{m_{1} - m_{2}}{2 c o s \frac{θ}{2}}, \frac{n_{1} - n_{2}}{2 c o s \frac{θ}{2}}

SOLUTION

Solution : B

Let OA and OB be two lines with d.c’s $l_{1}, m_{1}, n_{1}$ and $l_{2}, m_{2}, n_{2},$ . Let OA = OB = 1. Then, the coordinates of A and B are $(l_{1}, m_{1}, n_{1})$ and $(l_{2}, m_{2}, n_{2})$ , respectively. Let OC be the bisector of $∠ A O B$ .Then, C is the mid point of AB and so its coordinates are $(\frac{l_{1} + l_{2}}{2}, \frac{m_{1} + m_{2}}{2}, \frac{n_{1} + n_{2}}{2})$ .
$∴$ d.r's of OC are $\frac{l_{1} + l_{2}}{2}, \frac{m_{1} + m_{2}}{2}, \frac{n_{1} + n_{2}}{2}$
We have, $O C = \sqrt{{(\frac{l_{1} + l_{2}}{2})}^{2} + {(\frac{m_{1} + m_{2}}{2})}^{2} + {(\frac{n_{1} + n_{2}}{2})}^{2}} = \frac{1}{2} \sqrt{(l_{1}^{2} + m_{1}^{2} + n_{1}^{2}) + (l_{2}^{2} + m_{2}^{2} + n_{2}^{2}) + (l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2})} = \frac{1}{2} \sqrt{2 + 2 c o s θ} [Q c o s θ = l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2}] = \frac{1}{2} \sqrt{2 (1 + c o s θ)} = c o s (\frac{θ}{2})$

$∴$ d.c's of OC are
$\frac{l_{1} + l_{2}}{2 (O C)}, \frac{m_{1} + m_{2}}{2 (O C)}, \frac{n_{1} + n_{2}}{2 (O C)}$

Question 5

The projection of any line on co-ordinate axes be respectively 3, 4, 5 then its length is

A. 12

B. 50

5 \sqrt{2}

D. 25

SOLUTION

Solution : C

Let the line segment be AB, then as given
$A B c o s α = 3, A B c o s β = 4, A B c o s γ = 5 \Rightarrow A B^{2} (c o s^{2} α + c o s^{2} β + c o s^{2} γ) = 3^{2} + 4^{2} + 5^{2} A B = \sqrt{9 + 16 + 25} = 5 \sqrt{2}$
where $α, β$ and $γ$ are the angles made by the line with the axes.

Question 6

A variable plane which remains at a constant distance p from the origin cuts the coordinate axes in A, B, C. The locus of the centroid of the tetrahedron OABC is $y^{2} z^{2} + z^{2} x^{2} + x^{2} y^{2} = k x^{2} y^{2} z^{2}$ , where k is equal to

9 p^{2}

\frac{9}{p^{2}}

\frac{7}{p^{2}}

\frac{16}{p^{2}}

SOLUTION

Solution : D

Let equation of plane is
lx+my+nz=p
or $\frac{x}{(\frac{p}{l})} + \frac{y}{(\frac{p}{m})} + \frac{z}{(\frac{p}{n})} = 1$
Coordinates of A, B, C are $(\frac{p}{l}, 0, 0) (0, \frac{p}{m}, 0)$ and
$(0, 0, \frac{p}{n})$ respectively.
$∴$ Centroid of OABC is $(\frac{p}{4 l}, \frac{p}{4 m}, \frac{p}{4 n})$
$x_{1} = \frac{p}{4 l}, y_{1} = \frac{p}{4 m}, z_{1} = \frac{p}{4 n} ∵ l^{2} + m^{2} + n^{2} = 1 \Rightarrow \frac{p^{2}}{16 x_{1}^{2}} + \frac{p^{2}}{16 y_{1}^{2}} + \frac{p^{2}}{16 z_{1}^{2}} = 1$
or $x_{1}^{2} y_{1}^{2} + y_{1}^{2} z_{1}^{2} + z_{1}^{2} x_{1}^{2} = 16 / p^{2} x_{1}^{2} y_{1}^{2} z_{1}^{2}$
$∴$ Locus is $x^{2} y^{2} + y^{2} z^{2} + z^{2} x^{2} = \frac{16}{p^{2}} x^{2} y^{2} z^{2}$
$∴ k = \frac{16}{p^{2}}$

Question 7

The line joining the points (1, 1, 2) and (3, -2, 1) meets the plane 3x + 2y + z = 6 at the point

A. (1, 1, 2)

B. (3, -2, 1)

C. (2, -3, 1)

D. (3, 2, 1)

SOLUTION

Solution : B

The straight line joining the points (1, 1, 2) and (3, -2, 1) is $\frac{x - 1}{2} = \frac{y - 1}{- 3} = \frac{z - 2}{- 1} = r$ (say)
$∴$ Point is (2r + 1, 1 -3r, 2 –r )
Which lies on 3x + 2y + z = 6
$∴ 3 (2 r + 1) + 2 (1 - 3 r) + 2 - r = 6$
$∴ r = 1$
Required point is (3, -2, 1).

Question 8

The equation of the plane passing through the line of intersection of the planes 3x-y-4z=0 and x+3y+6=0 whose distance from the origin is 1, is

A. x -2y -2z-3 =0, 2x+y -2z+3 =0

B. x -2y+2z-3 =0, 2x+y+2z+3 =0

C. x+2y-2z-3=0, 2x-y-2z+3 =0

D. None of these

SOLUTION

Solution : A

Equation of planes passing through intersection of the planes 3x-y-4z=0 and x+3y+6=0 is, $(3 x - y - 4 z) + λ (x + 3 y + 6) = 0 . . . . . (i)$
Given, distance of plane (i) from origin is 1.
$∴ \frac{6 λ}{\sqrt{(3 + λ)^{2} + (3 λ - 1)^{2} + 4^{2}}} = 1$
or $36 λ^{2} = 10 λ^{2} + 26$ or $λ = \pm 1$
Put the value of $λ$ in (i),
$∴ (3 x - y - 4 z) \pm (x + 3 y + 6) = 0$
or 4x+2y -4z+6=0 i.e 2x+y-2z+3 =0
and 2x-4y-4z-6=0
Thus the required planes are x-2y-2z-3=0 and 2x+y-2z+3=0.

Question 9

The equation of the plane passing through the points (1,-3,-2) and perpendicular to planes x +2y+2z=5 and 3x+3y+2z=8, is

A. 2x -4y +3z-8 =0

B. 2x-4y-3z+8 =0

C. 2x+4y+3z+8 =0

D. x-3y+z-5=0

SOLUTION

Solution : A

l+2m+2n=0, 3l+3m+2n=0, $l^{2} + m^{2} + n^{2} = 1$ , we get l, m, n from these equations and then putting the values in l(x-1)+m(y+3) +n(z+2)=0, we get the required result.
Trick: Checking conversely,
2(x)-4(y)+3(z)-8 =0,
So, it passes through given point.
1(2)+2(-4)+2(3) =0,
So, it is perpendicular to x+2y+2z=5.
3(2)+3(-4)+2(3)=0,
So, it is perpendicular to 3x+3y +2z=8.

Question 10

The distance between two points P and Q is d and the length of their projections of PQ on the coordinate planes are $d_{1}, d_{2}, d_{3}$ . Then $d_{1}^{2} + d_{2}^{2} + d_{3}^{2} = k d^{2}$ , where k is

A. 1

B. 5

C. 3

D. 2

SOLUTION

Solution : D

Here, $d_{1} = d c o s (90^{\circ} - α)$
$d_{2} = d c o s (90^{\circ} - β)$
and $d_{3} = d c o s (90^{\circ} - γ)$
$∴ d_{1} = d s i n α$
$d_{2} = d s i n β$
and $d_{3} = d s i n γ$
$∴ d_{1}^{2} + d_{2}^{2} + d_{3}^{2} = k d^{2}$
$\Rightarrow d^{2} (s i n^{2} α + s i n^{2} β + s i n^{2} γ) = k (d)^{2}$
$∴ k = 2$

Free Three Dimensional Geometry 02 Practice Test - 12th Grade - Commerce

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

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Determinants 02
Application of Derivatives 03
Continuity and Differentiability 02