Three Dimensional Geometry Test

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Three Dimensional Geometry Test - 37

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Question 1
1 / -0

A plane passing through $$(-1, 2, 3)$$ and whose normal makes equal angle with the coordinate axes is

A
$$x + y + z + 4 = 0$$

B
$$x - y + z + 4 = 0$$

C
$$x + y + z - 4 = 0$$

D
$$x + y + z = 0$$

Solution

Since normal makes equal angles with coordinate axis.
So, it intercept with all the axis will be same. So equation of plane will be
$$\dfrac{x}{a}+\dfrac{y}{a}+\dfrac{z}{a}=1$$
$$\Rightarrow x+y+z=a$$
Now, it passes through $$(-1, 2, 3)$$, so
$$-1+2+3=a$$
$$\Rightarrow a=4$$
$$\Rightarrow x+y+z-4=0$$
Question 2
1 / -0

The direction cosine of a line which is perpendicular to both the lines whose direction ratios are $$1, 2, 2$$ and $$0, 2, 1$$ are

A
$$\displaystyle \frac { -2 }{ 3 } ,\frac { 1 }{ 3 } ,\frac { 2 }{ 3 } $$

B
$$\displaystyle \frac { 2 }{ 3 } ,\frac { -1 }{ 3 } ,\frac { 2 }{ 3 } $$

C
$$\displaystyle \frac { 2 }{ 3 } ,\frac { 1 }{ 3 } ,\frac { -2 }{ 3 } $$

D
$$\displaystyle \frac { 2 }{ 3 } ,\frac { -1 }{ 3 } ,\frac { -2 }{ 3 } $$

Solution

Since we need the direction cosines of a line perpendicular to both the lines, having direction ratios $$1,2,2$$ and $$0,2,1$$; we first find the direction ratios of the required line by finding the cross product of the given ratios.
It would result in $$(\hat{i} + 2\hat{j} + 2\hat{k}) \times (2\hat{j} + \hat{k})$$, i.e. $$2\hat{k} - \hat{j} + 0 - 2\hat{i} + 4\hat{i} + 0 $$
$$= 2\hat{i} - \hat{j} + 2\hat{k}$$

Thus, the direction ratios become $$2, -1, 2$$

And the corresponding direction cosines become $$\dfrac{2}{3}, \dfrac{-1}{3}, \dfrac{2}{3}$$
Question 3
1 / -0

Three district points A, B and C with p.v.s. and $$\displaystyle \vec { a } ,\vec { b } $$ and $$\displaystyle \vec { c } $$ respectively are collinear if there exist non-zero scalars x, y, z such that

A
$$\displaystyle x\vec { a } +y\vec { b } +z\vec { c } =0$$ and $$\displaystyle x+y+z=0$$

B
$$\displaystyle x\vec { a } +y\vec { b } +z\vec { c } $$ and $$\displaystyle x+y+z\neq 0$$

C
$$\displaystyle x\vec { a } +y\vec { b } +z\vec { c } \neq 0$$ and $$\displaystyle x+y+z=0$$

D
$$\displaystyle x\vec { a } +y\vec { b } +z\vec { c } =3$$ and $$\displaystyle x+y+z\neq 0$$

Solution

If three points are collinear, we can see that as one point dividing the line joining the other two points either internally or externally in some ratio.

So, $$x\vec {a} + y\vec {b} = (x + y)\vec{c}$$ when $$\vec{c}$$ divides the line joining $$\overline{a}$$ and $$\overline{b}$$ in the ratio $$y : x$$

This can also be written as $$x\vec {a} + y\vec {b} - (x + y)\vec {c} = 0$$

If $$z = - (x + y)$$, we get $$x + y + z = 0$$ and $$x\vec {a} + y\vec {b} + z\vec {c} = 0$$
Question 4
1 / -0

The direction ratios of two lines AB, AC are 1, -1, -1 and 2, -1, 1. The direction ratios of the normal to the plane ABC are

A
$$2, 3, -1$$

B
$$2, 2, 1$$

C
$$3, 2, -1$$

D
$$-1, 2, 3$$

Solution

Given : Direction ratios of $$AB$$ and $$AC$$ are $$1,-1,-1$$ and $$2,-1,1$$
i.e. $$\vec{AB}=\hat{i}-\hat{j}-\hat{k}$$ and $$\vec{AC}=2\hat{i}-\hat{j}+\hat{k}$$

$$\therefore$$ Normal vector to the plane ABC is given by
$$=\vec{AC}\times \vec{AB}$$
$$=(2\hat{i}-\hat{j}+\hat{k}) \times (\hat{i}-\hat{j}-\hat{k})$$
$$=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 1 \\ 1 & -1 & -1 \end{vmatrix}$$

$$=2\hat{i}+3\hat{j}-\hat{k}$$
Hence, the direction ratios of normal are $$2,3,-1$$
Question 5
1 / -0

If $$A(3, 4, 5), B(4, 6, 3), C(-1, 2, 4)$$ and $$D(1, 0, 5)$$ are such that the angle between the lines $$\overline{DC}$$ and $$\overline{AB}$$ is $$\theta$$ then $$cos\,\theta =$$

A
$$\dfrac{7}{9}$$

B
$$\dfrac{2}{9}$$

C
$$\dfrac{4}{9}$$

D
$$\dfrac{5}{9}$$

Solution

Given, $$A(3,4,5), B(4,6,3), C(-1,2,4), D(1,0,5)$$

$$\vec{DC}=-2i+2j-k$$...(i)
$$\vec{AB}=i+2j-2k$$ ...(ii)

Hence, $$DC.AB\cos\theta=\vec{DC}.\vec{AB}$$

$$9\cos\theta=-2+4+2$$
$$\cos\theta=\dfrac{4}{9}$$
Question 6
1 / -0

If a line makes angles $$\alpha, \beta, \gamma$$ with the coordinate axes, then the value of $$\cos 2\alpha + \cos 2\beta + \cos 2\gamma$$ is

A
$$3$$

B
$$-2$$

C
$$2$$

D
$$-1$$

Solution

$$\cos 2\alpha + \cos 2\beta + \cos 2\gamma$$
$$= 2(\cos^{2} \alpha + \cos^{2}\beta + \cos^{2}\gamma) - 3$$
$$= 2 - 3 = -1$$
$$(\because \cos^{2}\alpha + \cos^{2}\beta + \cos^{2}\gamma = 1)$$
Question 7
1 / -0

The angle between two diagonals of a cube is.

A
$$30^o$$

B
$$45^o$$

C
$$cos^{-1}(\frac{1}{3})$$

D
$$cos^{-1}(\frac{1}{\sqrt{3}})$$

Solution

We know that $$cos\theta = \dfrac{\left | \vec{u}.\vec{v} \right |}{\left | \vec{u} \right |\left | \vec{v} \right |}$$
Consider a cube with one vertex at the origin.
Then one diagonal is from $$(0, 0, 0)$$ to $$(1, 1, 1)$$: $$\left | \vec{u} \right |=<1, 1, 1>$$
Another diagonal is from $$(1, 0, 0)$$ to $$(0, 1, 1)$$: $$\left | \vec{v} \right |=<1, -1, -1>$$
Therefore, $$cos\theta = \dfrac{\left | <1, 1,1 >.<1, -1, -1> \right |}{\left | \sqrt{1^1+1^2+1^2} \right |\left | \sqrt{1^2+(-1)^2+(-1)^2} \right |}$$
$$\Rightarrow \dfrac{\left | 1-1-1 \right |}{\sqrt{3}\sqrt{3}}\Rightarrow \dfrac{1}{3}$$
Therefore, $$\theta=cos^{-1}\left ( \frac{1}{3} \right )$$
Question 8
1 / -0

If $$\cos { \alpha } ,\cos { \beta } ,\cos { \gamma } $$ are the direction cosines of a vector $$\vec { a } $$, then $$\cos { 2\alpha } +\cos { 2\beta } +\cos { 2\gamma } $$ is equal to

A
$$2$$

B
$$3$$

C
$$-1$$

D
$$0$$

Solution

We know that $$\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1$$ ....(1)
Hence $$\cos2\alpha+\cos2\beta+\cos2\gamma=(2\cos^2\alpha-1)+(2\cos^2\beta-1)+(2\cos^2\gamma-1)$$
$$=2(\cos^2\alpha+\cos^2\beta+\cos^2\gamma)-3=2\cdot 1-3=-1$$, using (1)
Question 9
1 / -0

The vector equation of the plane which is at a distance of $$\cfrac { 3 }{ \sqrt { 14 } } $$ from the origin and the normal from the origin is $$2\hat { i } -3\hat { j } +\hat { k } $$ is

A
$$\vec { r } .(2\hat { i } -3\hat { j } +\hat { k } )=3$$

B
$$\vec { r } .(\hat { i } +\hat { j } +\hat { k } )=9$$

C
$$\vec { r } .(\hat { i } +2\hat { j } )=3$$

D
$$\vec { r } .(2\hat { i } +\hat { k } )=3$$

Solution

The unit vector along the normal of the plane is given by
$$\vec{n}=\dfrac{2i-3j+k}{\sqrt{4+9+1}}=\dfrac{2i-3j+k}{\sqrt{14}}$$
Therefore, the vector equation of the plane with the above normal is
$$\vec{r}.(\dfrac{2i-3j+k}{\sqrt{14}})=d$$ where 'd' is the distance of the plane from the origin.
Now it is given that $$d=\dfrac{3}{\sqrt{14}}$$, hence the vector equation of the plane becomes
$$\vec{r}.(\dfrac{2i-3j+k}{\sqrt{14}})=\dfrac{3}{\sqrt{14}}$$ or

$$\vec{r}.(2i-3j+k)=3$$.
Question 10
1 / -0

If the angles made by a straight line with the coordinate axes are $$\alpha, \dfrac{\pi}{2}-\alpha, \beta$$ then $$\beta=$$

A
$$0$$

B
$$\dfrac{\pi}{6}$$

C
$$\dfrac{\pi}{2}$$

D
$$\pi$$

Solution

Given angles are $$\alpha, 90^0-\alpha, \beta$$

The summations of cosines of these angles has to be $$1$$.

So, $$\cos^2\alpha + \cos^2(90^0-\alpha)+\cos^2\beta=1$$
$$\cos^2\alpha+\sin^2\alpha+\cos^2\beta=1$$

$$\cos^2\beta=0$$

$$\beta=\dfrac{\pi}{2}$$

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Three Dimensional Geometry Test - 37

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Three Dimensional Geometry Test - 37

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Question 1

$$x + y + z + 4 = 0$$

$$x - y + z + 4 = 0$$

$$x + y + z - 4 = 0$$

$$x + y + z = 0$$

Solution

Question 2

$$\displaystyle \frac { -2 }{ 3 } ,\frac { 1 }{ 3 } ,\frac { 2 }{ 3 } $$

$$\displaystyle \frac { 2 }{ 3 } ,\frac { -1 }{ 3 } ,\frac { 2 }{ 3 } $$

$$\displaystyle \frac { 2 }{ 3 } ,\frac { 1 }{ 3 } ,\frac { -2 }{ 3 } $$

$$\displaystyle \frac { 2 }{ 3 } ,\frac { -1 }{ 3 } ,\frac { -2 }{ 3 } $$

Solution

Question 3

$$\displaystyle x\vec { a } +y\vec { b } +z\vec { c } =0$$ and $$\displaystyle x+y+z=0$$

$$\displaystyle x\vec { a } +y\vec { b } +z\vec { c } $$ and $$\displaystyle x+y+z\neq 0$$

$$\displaystyle x\vec { a } +y\vec { b } +z\vec { c } \neq 0$$ and $$\displaystyle x+y+z=0$$

$$\displaystyle x\vec { a } +y\vec { b } +z\vec { c } =3$$ and $$\displaystyle x+y+z\neq 0$$

Solution

Question 4

$$2, 3, -1$$

$$2, 2, 1$$

$$3, 2, -1$$

$$-1, 2, 3$$

Solution

Question 5

$$\dfrac{7}{9}$$

$$\dfrac{2}{9}$$

$$\dfrac{4}{9}$$

$$\dfrac{5}{9}$$

Solution

Question 6

$$3$$

$$-2$$

$$2$$

$$-1$$

Solution

Question 7

$$30^o$$

$$45^o$$

$$cos^{-1}(\frac{1}{3})$$

$$cos^{-1}(\frac{1}{\sqrt{3}})$$

Solution

Question 8

$$2$$

$$3$$

$$-1$$

$$0$$

Solution

Question 9

$$\vec { r } .(2\hat { i } -3\hat { j } +\hat { k } )=3$$

$$\vec { r } .(\hat { i } +\hat { j } +\hat { k } )=9$$

$$\vec { r } .(\hat { i } +2\hat { j } )=3$$

$$\vec { r } .(2\hat { i } +\hat { k } )=3$$

Solution

Question 10

$$0$$

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

$$\dfrac{\pi}{2}$$

$$\pi$$

Solution

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