Free Three Dimensional Geometry 03 Practice Test - 12th Grade - Commerce

Question 1

The plane $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$ meets the coordinate axes at A, B, C respectively. D and E are the mid-points of AB and AC respectively. Coordinates of the mid-point of DE are

(a, \frac{b}{4}, \frac{c}{4})

(\frac{a}{4}, b, \frac{c}{4})

(\frac{a}{4}, \frac{b}{4}, c)

(\frac{a}{2}, \frac{b}{4}, \frac{c}{4})

SOLUTION

Solution : D

$A (a, 0, 0), B (0, b, 0), C (0, 0, c), D (\frac{a}{2}, \frac{b}{2}, 0), E (\frac{a}{2}, 0, \frac{c}{2})$
So midpoint of DE is $(\frac{a}{2}, \frac{b}{4}, \frac{c}{4})$ .

Question 2

The ratio in which the plane 2x - 1 = 0 divides the line joining (-2,4,7) and (3, -5, 8) is

A. 2 : 3

B. 4 : 5

C. 7 : 8

D. 1 : 1

SOLUTION

Solution : D

Let the required ratio be k:1, then the coordinates of the point which divides the join of the points (-2, 4, 7) and (3, -5, 8) in this ratio are given by $(\frac{3 k - 2}{k + 1}, \frac{- 5 k + 4}{k + 1}, \frac{8 k + 7}{k + 1})$
As this point lies on the plane 2x - 1 = 0.
$(\frac{3 k - 2}{k + 1} = \frac{1}{2} \Rightarrow k = 1)$ and thus the required ratio as 1:1.

Question 3

If a plane passes through the point (1,1,1) and is perpendicular to the line $\frac{x - 1}{3} = \frac{y - 1}{0} = \frac{z - 1}{4}$ then its perpendicular distance from the origin is

\frac{3}{4}

\frac{4}{3}

\frac{7}{5}

1

SOLUTION

Solution : C

The d.r of the normal to the plane is $(3, 0, 4)$ The equation of the plane is 3x + 0y + 4z + d = 0 since it passes through (1, 1, 1) so; d = --7
Now distance of the plane $3 x + 4 z - 7 = 0$ from (0, 0, 0) is $\frac{7}{\sqrt{3^{2} + 4^{2}}} = \frac{7}{5}$ units

Question 4

If centroid of the tetrahedron , whose vertices are given by (0,0,0), (a, 2, 3),(1, b, 2) and (2, 1, c) be (1, 2, –1), then distance of P(a,b,c) from origin is equal to

\sqrt{107}

\sqrt{14}

\frac{\sqrt{107}}{\sqrt{14}}

10

SOLUTION

Solution : A

Centroid $\equiv (\frac{\sum x}{4}, \frac{\sum y}{4}, \frac{\sum z}{4}) = (1, 2, - 1)$
$\Rightarrow a = 1, b = 5, c = - 9; ∴ \sqrt{a^{2} + b^{2} + c^{2}} = \sqrt{107}$ .

Question 5

If the lines $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4}$ and $\frac{x - 3}{1} = \frac{y - k}{1} = \frac{z}{1}$ intersect, then k =

\frac{2}{9}

\frac{9}{2}

C. 0

D. 3

SOLUTION

Solution : B

Any point on $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4} = λ$ is,
$(2 λ + 1, 3 λ - 1, 4 λ + 1); λ \in R$
Any point on $\frac{x - 3}{1} = \frac{y - k}{2} = \frac{z}{1} = μ$ is,
$(μ + 3, 2 μ + k, μ); μ \in R$
the given lines intersect if and only if the system of equations (in $λ$ and $μ)$
$2 λ + 1 = μ + 3 . . . . (i)$
$3 λ - 1 = 2 μ + k . . . . . (i i)$
$4 λ + 1 = μ . . . . (i i i)$
has a unique solution.
Solving (i) and (iii), we get $λ = \frac{- 3}{2}, μ = - 5$
From (ii), we get $\frac{- 9}{2} - 1 = - 10 + k \Rightarrow k = \frac{9}{2}$ .

Question 6

If a line makes the angle $α, β, γ$ with three dimensional co-ordinate axes respectively, then $c o s 2 α + c o s 2 β + c o s 2 γ$

A. -2

B. -1

C. 1

D. 2

SOLUTION

Solution : B

$c o s 2 α + c o s 2 β + c o s 2 γ = 2 c o s^{2} α - 1 + 2 {cos}^{2} β - 1 + 2 c o s^{2} γ - 1 = 2 (c o s^{2} α + c o s^{2} β + c o s^{2} γ) - 3 = 2 - 3 = - 1$

Question 7

Perpendicular distance of the point (3, 4, 5) from the y-axis, is

\sqrt{34}

\sqrt{41}

C. 4

D. 5

SOLUTION

Solution : A

Perpendicular distance of (x,y,z) from y axis is given by $= \sqrt{x^{2} + z^{2}}$ .
=> Required distance $= \sqrt{3^{2} + 5^{2}} = \sqrt{34}$ .

Question 8

The perpendicular distance of the origin form the plane which makes intercepts 12, 3 and 4 on x, y, z axes respectively, is

A. 13

B. 11

C. 17

\frac{6 \sqrt{2}}{\sqrt{1} 3}

SOLUTION

Solution : D

Let equation of plane be lx + my + nz = p, where p is the perpendicular distance of the plane from origin and (l,m,n) are the direction cosines of the normal to the plane
Or $\frac{x}{(\frac{p}{l})} + \frac{y}{(\frac{p}{m})} + \frac{z}{(\frac{p}{n})} = 1$
According to question,
$\frac{p}{l} = 12, \frac{p}{m} = 3, \frac{p}{n} = 4$
or $\frac{p}{12} = l, \frac{p}{3} = m, \frac{p}{4} = n$
or $\frac{p^{2}}{144} + \frac{p^{2}}{9} + \frac{p^{2}}{16} = l^{2} + m^{2} + n^{2} = 1$
$\Rightarrow p^{2} (\frac{1}{144} + \frac{1}{9} + \frac{1}{16}) = 1 p^{2} (\frac{1 + 16 + 9}{144}) = 1$
Or $p^{2} = \frac{144}{26} = \frac{72}{13}$
$∴ p = \frac{6 \sqrt{2}}{\sqrt{13}}$

Question 9

The equation of the plane passing through the points (2, –4, 0), (3, –4, 5) and parallel to the line 2x = 3y = 4z is :

A. 125x – 90y – 79z = 340

B. 32x – 21y – 36z = 85

C. 73x + 61y – 22z = 85

D. 29x – 27y – 22z = 85

SOLUTION

Solution : D

Equation of the plane through the point (2, –1, 0) is
a(x – 2) + b(y + 1) + c(z – 0) = 0 . . . (1)
It passes through the point (3, –4, 5)
$∴$ a – 3b + 5c = 0 . . . (2)
The plane (1) is parallel to the line
2x = 3y = 4z i.e., $\frac{x}{6} = \frac{y}{4} = \frac{z}{3} . . . (3)$
$∴$ Normal to the plane (1) is perpendicular to the line (3).
$∴$ 6a + 4b + 3c = 0 . . .(4)
From (2) and (4), we get
$\frac{a}{- 29} = \frac{b}{27} = \frac{c}{22}$
$∴$ Equation of the required plane is
– 29(x – 2) + 27(y + 1) + 22(z) = 0
$\Rightarrow - 29 x + 27 y + 22 z + 85 = 0$
$\Rightarrow 29 x - 27 y - 22 z = 85$

Question 10

The distance of the point (1, –2, 3) from the plane x – y + z = 5 measured parallel to the line $(\frac{1}{2}) x = (\frac{1}{3}) y = (\frac{- 1}{6}) z$ is :

A. 1

B. 2

\frac{1}{2}

D. 4

SOLUTION

Solution : A

Line parallel to the line $\frac{x}{2} = \frac{y}{2} = \frac{z}{- 6}$ and passing through (1,-2, 3) is
$\frac{x - 1}{2} = \frac{y + 2}{3} = \frac{z - 3}{- 6} = r$
A = (2r + 1, 3r – 2, – 6r + 3) is a point on this line.
A lies in the plane x – y + z = 5
$∴ 2 r + 1 - 3 r + 2 - 6 r + 3 = 5 \Rightarrow r = \frac{1}{7} ∴ A = (\frac{9}{7}, \frac{- 11}{7}, \frac{15}{7})$
Distance of A from (1, –2, 3) is
$\sqrt{{(\frac{9}{7} - 1)}^{2} + {(\frac{- 11}{7} + 2)}^{2} + {(\frac{15}{7} - 3)}^{2}} = \frac{1}{7} \sqrt{4 + 9 + 36} = \frac{7}{7} = 1$

Free Three Dimensional Geometry 03 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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