Three Dimensional Geometry Test

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Three Dimensional Geometry Test - 59

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Question 1
1 / -0

$$L_{1}$$ and $$L_{2}$$ are two lines whose vector equations are
$$L_{1} = \vec {r} = \lambda [(\cos \theta + \sqrt {3})\hat {i} + (\sqrt {2}\sin \theta)\hat {j} + (\cos \theta - \sqrt {3})\hat {k}]$$
$$L_{2} = \vec {r} = \mu (a\hat {i} + b\hat {j} + c\hat {k})$$, where $$\lambda$$ and $$\mu$$ are scalars and $$\alpha$$ is the acute angle between $$L_{1}$$ and $$L_{2}$$. If the angle $$'\alpha'$$ is independent of $$\theta$$ then the value of $$'\alpha'$$ is

A
$$\dfrac {\pi}{6}$$

B
$$\dfrac {\pi}{4}$$

C
$$\dfrac {\pi}{3}$$

D
None of these

Solution
Question 2
1 / -0

The equation of the line parallel to $$\cfrac { x-3 }{ 1 } =\cfrac { y+3 }{ 5 } =\cfrac { 2z-5 }{ 3 } $$ and passing through the point $$(1,3,5)$$ in vector form, is:

A
$$\vec { r } =\left( \vec { i } +5\vec { j } +3\vec { k } \right) +t\left( \vec { i } +3\vec { j } +5\vec { k } \right) $$

B
$$\vec { r } =\left( \vec { i } +3\vec { j } +5\vec { k } \right) +t\left( \vec { i } +5\vec { j } +3\vec { k } \right) $$

C
$$\vec { r } =\left( \vec { i } +5\vec { j } +\cfrac { 3 }{ 2 } \vec { k } \right) +t\left( \vec { i } +3\vec { j } +5\vec { k } \right) $$

D
$$\vec { r } =\left( \vec { i } +3\vec { j } +5\vec { k } \right) +t\left( \vec { i } +5\vec { j } +\cfrac { 3 }{ 2 } \vec { k } \right) $$

Solution

We have to find the equation of the line parallel to $$\dfrac{x-3}{1}=\dfrac{y+3}{5}=\dfrac{2z-5}{3}$$ and passing through the point $$(1,3,5)$$ vector form.

We know that the equation of the line passing through the point with position vector $$\vec{a}$$ and parallel to the vector $$\vec{b}$$ is $$\vec{r}=\vec{a}+t\vec{b}$$

Consider $$\dfrac{x-3}{1}=\dfrac{y+3}{5}=\dfrac{2z-5}{3}$$

Rewriting we get $$\dfrac{x-3}{1}=\dfrac{y+3}{5}=\dfrac{z-\dfrac{5}{2}}{\dfrac{3}{2}}$$

Thus vector representation is $$\vec{b}=\vec{i}+5\vec{j}+\dfrac{3}{2}\vec{k}$$

Since line passes through $$(1,3,5)$$, $$\vec{a}=\vec{i}+3\vec{j}+5\vec{k}$$

Hence the required equation of the line is $$\vec{r}=(\vec{i}+3\vec{j}+5\vec{k})+t\left(\vec{i}+5\vec{j}+\dfrac{3}{2}\vec{k}\right)$$
Question 3
1 / -0

If points $$P\left( 4,5,x \right) ,Q\left( 3,y,4 \right) $$ and $$ R\left( 5,8,0 \right) $$ are colinear, then the value of $$x+y$$ is

A
$$-4$$

B
$$3$$

C
$$5$$

D
$$4$$

Solution

$$\vec{PR}=(5-4)\hat{i}+(8-5)\hat{j}+(0-x)\hat{k}$$

$$\implies \vec{PR}=\hat{i}+3\hat{j}-x\hat{k}$$

Again, $$\vec{QR}=(5-3)\hat{i}+(8-y)\hat{j}+(0-4)\hat{k}$$

$$\implies \vec{QR}=2\hat{i}+(8-y)\hat{j}-4\hat{k}$$

Since, P,Q and R are co-linear points, hence

$$\dfrac{1}{2}=\dfrac{3}{8-y}=\dfrac{-x}{-4}$$

On Solving, we get:-

$$x=2,y=2$$

$$\implies x+y=4$$

Hence, answer is option-(D).
Question 4
1 / -0

If $$\alpha,\beta,\gamma\in[0,2\pi]$$, then the sum of all possible values of $$\alpha, \beta,\gamma$$ if $$\sin \alpha=-\dfrac{1}{\sqrt{2}}$$, $$\cos \beta=-\dfrac{1}{2}$$, $$\tan \gamma=-\sqrt{3}$$, is

A
$$\dfrac{22\pi}{3}$$

B
$$\dfrac{21\pi}{3}$$

C
$$\dfrac{20\pi}{3}$$

D
$$8\pi$$

Solution

$$\sin\alpha = \dfrac{-1}{\sqrt2}$$

(-ve) value lies on III, IV quadrant for sin

$$\therefore \alpha = \pi + \dfrac{\pi}{4}, 2\pi - \dfrac{\pi}{4}$$

$$= \dfrac{5\pi}{4}, \dfrac{7\pi}{4}$$

$$\cos\beta = \dfrac{-1}{2}$$

(-ve) value lies on II, III quadrant for cosine

$$ \beta = \pi - \dfrac{\pi}{3}, \pi + \dfrac{\pi}{3}$$

$$= \dfrac{2\pi}{3}, \dfrac{4\pi}{3}$$

$$\tan\gamma = -\sqrt3$$

(-ve) value lies on II, IV quadrant for tan : $$\gamma = -\sqrt3$$

$$ \gamma = \pi - \dfrac{\pi}{3}, 2\pi - \dfrac{\pi}{3}$$

The sum of all values $$(\alpha + \beta + \gamma)= \dfrac{2\pi}{3}, \dfrac{5\pi}{3}$$

$$= \dfrac{5\pi}{4} + \dfrac{7\pi}{4} + \dfrac{2\pi}{3} + \dfrac{4\pi}{3} + \dfrac{2\pi}{3} + \dfrac{7\pi}{3}$$

$$= 3\pi + 2\pi + \dfrac{7\pi}{3}$$

$$= 5\pi + \dfrac{7\pi}{3}$$

$$= \dfrac{22\pi}{3}$$
Question 5
1 / -0

In $$\Delta ABC, |\bar{CB}| = a, |\bar{CA}| = b, |\bar{AB}| = c$$. $$CD$$ is median through the vertex $$C$$. Then $$\bar{CA}.\bar{CD}$$ equals.

A
$$\dfrac{1}{4}(3a^2 + b^2 - c^2)$$

B
$$\dfrac{1}{4}(a^2 + 3b^2 - c^2)$$

C
$$\dfrac{1}{4}(a^2 + b^2 - 3c^2)$$

D
$$\dfrac{1}{4}(-3a^2 + b^2 - c^2)$$

Solution

$$\overrightarrow{CD}=\overrightarrow{CA}+\overrightarrow{AD}$$
$$\overrightarrow{AD}=\dfrac{1}{2}\overrightarrow{AB}$$
$$\overrightarrow{CD}=\overrightarrow{CA}+\dfrac{1}{2}\overrightarrow{AB}$$
$$\overrightarrow{CD}.\overrightarrow{CA}=\left(\overrightarrow{CA}+\dfrac{1}{2}\overrightarrow{AB}\right)\overrightarrow{CA}$$
$$|\overrightarrow{CA}|^2+\dfrac{1}{2}\overrightarrow{AB}.\overrightarrow{CA}$$
$$b^2+\dfrac{1}{2}\overrightarrow{AB}.\overrightarrow{CA}$$ $$(|\overrightarrow{CA}|=b)$$
$$=\overrightarrow{b}+\dfrac{1}{2}|\overrightarrow{AB}||\overrightarrow{CA}|\cos (\pi-A)$$
$$b^2-\dfrac{1}{2}bc\left(\dfrac{b^2+c^2-a^2}{2ac}\right)$$
$$b^2-\dfrac{b^2}{4}-\dfrac{c^2}{4}+\dfrac{a^2}{4}$$
$$\dfrac{3b^2}{4}-\dfrac{c^2}{4}+\dfrac{a^2}{4}$$
$$\dfrac{1}{4}(a^2+3b^2-c^2)$$
Question 6
1 / -0

The equation of plane passing through a point $$A(2, - 1, 3)$$ and parallel to the vectors $$a= (3, 0, - 1)$$ and $$b=(- 3, 2, 2)$$ is:

A
$$2x - 3y + 6z - 25 = 0$$

B
$$2x - 3y + 6z + 25 = 0$$

C
$$3x - 2y + 6z - 25 = 0$$

D
$$3x - 2y + 6z + 25 = 0$$

Solution

Let the DR of the normal of required plane be $$<a, b, c>$$
Since plane is parallel to $$(3, 0, -1)$$
$$\therefore$$ normal must be perpendicular
$$\therefore 3a + 0b - c = 0$$ ... $$(1)$$
Also, it is parallel to $$(-3, 2, 2)$$
$$\therefore -3a + 2b + 2c = 0$$ ... $$(2)$$
From $$(1)$$ and $$(2)$$
$$\dfrac{a}{2} = \dfrac{-b}{6 - 3} = \dfrac{c}{6}$$
$$a = 2, b = -3, c = 6$$
It passes through $$(2, -1, 3)$$
$$2(x - 2) - 3(y + 1) + 6(z - 3) = 0$$
$$\Rightarrow 2x - 4 - 3y - 3 + 6z - 18 = 0$$
$$\Rightarrow 2x - 3y + 6z - 25 = 0$$
Question 7
1 / -0

If a = 4i + 3j and b be two vectors perpendicular to each other on the xy- plane. Then, a vector in the same plane having projections 1 and 2 along a and b respectively, is

A
i + 2j

B
2i - j

C
2i + j

D
None of these

Solution

Let $$\vec c$$ be the required vector
Given that $$\vec a =4\hat i+3\hat j$$
Since, you have not given $$\vec b $$, it is assumed to be $$=-3\hat i+4\hat j$$
Projection of $$\vec c \theta n \vec a=\dfrac {\vec c.\vec a}{|\vec a|}=1$$
$$\boxed {\vec c.\vec a=5}$$
Probability of $$\vec c$$ on $$\vec b=\dfrac {\vec c.\vec b}{(\vec b)}=2, \vec c.\vec b=10$$
Let $$\vec c=n\hat i+y\hat j$$
$$\vec c.\vec b=100, -3x+y=10$$
$$x=\dfrac {-2}{5}, y=\dfrac {11}{5}$$
$$\boxed {\vec =\dfrac {-2}{5}\hat i+\dfrac {11}{5}\hat j}$$
Question 8
1 / -0

The equation of the plane passing through (1, -2, 4), (3, -4, 5) and perpendicular to yz-plane is.

A
2y + z = 0

B
y + 2y + 6 = 0

C
y + 2y - 6 = 0

D
3y + 2z - 2 = 0

Solution
Question 9
1 / -0

If a line makes angle $$90^{o},135^{o},45^{o}$$ with the $$X-$$,$$Y-$$ and $$Z-$$axes respectively, then its direction cosines are

A
$$0,\dfrac{1}{\sqrt{2}},-\dfrac{1}{\sqrt{2}}$$

B
$$0,-\dfrac{1}{\sqrt{2}},-\dfrac{1}{\sqrt{2}}$$

C
$$0,\dfrac{1}{\sqrt{2}},\dfrac{1}{\sqrt{2}}$$

D
$$0,-\dfrac{1}{\sqrt{2}},\dfrac{1}{\sqrt{2}}$$

Solution
Question 10
1 / -0

In the isosceles $$\triangle$$ABC, $$|AB| = |BC|=8$$ and point E divides AB internally in the ratio 1 : 3 then the cosine of angel between CE and CA is (where, |CA| = 12) ?

A
$$-\dfrac { \sqrt [ 3 ]{ 7 } }{ 8 }$$

B
$$\dfrac { \sqrt [ 3 ]{ 8 } }{ 17 }$$

C
$$\dfrac { \sqrt [ 3 ]{ 7 } }{ 8 }$$

D
$$-\dfrac { \sqrt [ 3 ]{ 8 } }{ 17 }$$

Solution