Three Dimensional Geometry Test

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Three Dimensional Geometry Test - 46

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Weekly Quiz Competition

Question 1
1 / -0

Let $$\overrightarrow{p}=3a{x}^{2}\hat{i}-2\left(x-1\right)\hat{j}$$ and $$\overrightarrow{q}=b\left(x-1\right)\hat{i}+x\hat{j}$$. If $$ab<0$$ then $$\overrightarrow{p}$$ and $$\overrightarrow{q}$$ are parallel for

A
atleast one $$x$$ is $$\left(0,1\right)$$

B
atleast one $$x$$ is $$\left(-1,0\right)$$

C
atleast one $$x$$ is $$\left(1,2\right)$$

D
none of these

Solution

$$\overrightarrow{p},\overrightarrow{q}$$ are parallel if $$\overrightarrow{p}\times \overrightarrow{q}=0$$ and $$\overrightarrow{p}\times \overrightarrow{q}$$
$$=\left|\begin{matrix}\hat{i}& \hat{j} & \hat{k}\\ 3a{x}^{2} &-2\left(x-1\right)&0 \\ b\left(x-1\right) & x & 0 \end{matrix}\right|$$
$$=\hat{i}\left(0-0\right)-\hat{j}\left(0-0\right)+\hat{k}\left(3a{x}^{3}+2b{\left(x-1\right)}^{2}\right)$$
$$=\hat{k}\left(3a{x}^{3}+2b{\left(x-1\right)}^{2}\right)$$
$$\left|\overrightarrow{p}\times\overrightarrow{q}\right|=3a{x}^{3}+2{\left(x-1\right)}^{2}b=f\left(x\right)$$
$$f\left(0\right)=2b$$ , $$f\left(1\right)=3a$$
$$f\left(0\right).f\left(1\right)=2b\times 3a=6ab<0$$ (given)
Question 2
1 / -0

If $$\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$$ are non-coplanar and $$\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\alpha\overrightarrow{d}$$ , $$\overrightarrow{b}+\overrightarrow{c}+\overrightarrow{d}=\beta\overrightarrow{a}$$ then $$\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}+\overrightarrow{d}=$$

A
$$0$$

B
$$\alpha\overrightarrow{a}$$

C
$$\beta\overrightarrow{b} $$

D
$$\left(\alpha+\beta\right)\overrightarrow{c}$$

Solution

$$\overrightarrow{a} +\overrightarrow{b} +\overrightarrow{c} =\alpha\overrightarrow{d} \left(given\right)$$
$$\overrightarrow{b} +\overrightarrow{c} +\overrightarrow{d} =\beta\overrightarrow{a} \left(given\right)$$
Eliminating $$\overrightarrow{d}$$
$$\Rightarrow \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\alpha\overrightarrow{d} $$
$$\Rightarrow \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\alpha\left(\beta\overrightarrow{a}-\overrightarrow{b}-\overrightarrow{c}\right)$$
$$\Rightarrow \left(1-\alpha\beta\right)\overrightarrow{a}+\left(1+\alpha\right)\overrightarrow{b}+\left(1+\alpha\right)\overrightarrow{c}=0$$
since $$\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$$ are non-coplanar
$$\Rightarrow 1-\alpha\beta=0, 1+\alpha=0$$
$$\Rightarrow \alpha\beta=1, \alpha=-1$$
$$\Rightarrow \alpha=-1 $$
$$\therefore \beta=-1$$
$$\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\alpha\overrightarrow{d}=-\overrightarrow{d}$$
$$\therefore \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}+\overrightarrow{d}=0$$
Question 3
1 / -0

If the angle between the line $$ x = \dfrac { y - 1 } { 2 } = \dfrac { z - 3 } { \lambda } $$ and the plane $$ x + 2 y + 3 z = 4 \text { is } \cos ^ { - 1 } \left( \sqrt { \frac { 5 } { 14 } } \right) $$, then $$ \lambda $$ equals:-

A
$$ \frac{2}{5} $$

B
$$ \frac{5}{3} $$

C
$$ \frac{2}{3} $$

D
$$ \frac{3}{2} $$

Solution

We have,
$$x = \dfrac{{y - 1}}{2} = \dfrac{{z - 3}}{\lambda }\,\,\,\,\quad x + 2y + 3z = 4$$

$$ = {\cos ^{ - 1}}\left( {\dfrac{{\sqrt 5 }}{{14}}} \right)\,\,\, \quad \lambda = ?$$

$$ = \cos = \dfrac{{\sqrt 5 }}{{14}}$$

$$\theta = {\sin ^{ - 1}}\sqrt {1 - \dfrac{5}{{14}}} = {\sin ^{ - 1}}\left( {\dfrac{{\sqrt 9 }}{{14}}} \right)$$

$$ \sin \theta =\sqrt { \dfrac { 9 }{ { 14 } } }$$

$$ \Rightarrow\dfrac { { 1+4+3\lambda } }{ { \sqrt { 5+{ \lambda ^{ 2 } } } \sqrt { 14 } } } = \dfrac { 3 }{ { \sqrt { 14 } } } $$..................obtained from the cross product relation of two normal vectors of the line and the plane.

$$\Rightarrow \dfrac { { 5+3\lambda } }{ { \sqrt { 5+{ \lambda ^{ 2 } } } \sqrt { 14 } } } =\dfrac{3}{\sqrt {14}} $$

$$ \Rightarrow 25+9{ \lambda ^{ 2 } }+30\lambda =9\left( { 5+\lambda^2 } \right) $$

$$ \Rightarrow 25 + 30\lambda =45 $$

$$20 = 30\lambda $$

$$\lambda = \dfrac{2}{3}$$
Then,

Option $$C$$ is correct answer.
Question 4
1 / -0

$$A=(-1, 2, -3), B=(5, 0, -6), C=(0, 4, -1)$$ are the vertices of a triangle. The d.c's of the internal bisector of $$\angle$$BAC are?

A
$$\left(\dfrac{25}{\sqrt{714}}, \dfrac{-8}{\sqrt{714}}, \dfrac{-5}{\sqrt{714}}\right)$$

B
$$\left(\dfrac{5}{\sqrt{74}}, \dfrac{6}{\sqrt{74}}, \dfrac{8}{\sqrt{74}}\right)$$

C
$$\left(\dfrac{25}{\sqrt{714}}, \dfrac{8}{\sqrt{714}}, \dfrac{5}{\sqrt{714}}\right)$$

D
$$\left(\dfrac{-5}{\sqrt{74}}, \dfrac{6}{\sqrt{74}}, \dfrac{-8}{\sqrt{74}}\right)$$

Solution

Here,
$$\begin{array}{l} AB=\sqrt { { { \left( { 5+1 } \right) }^{ 2 } }+{ { \left( { 0-2 } \right) }^{ 2 } }+{ { \left( { -6+3 } \right) }^{ 2 } } } =\sqrt { 49 } =7 \\ AC=\sqrt { { { \left( { 0+1 } \right) }^{ 2 } }+{ { \left( { 4-2 } \right) }^{ 2 } }+{ { \left( { 1+3 } \right) }^{ 2 } } } =\sqrt { 9 } =3 \end{array}$$
by geometry , the bisector of$$\angle BAC$$
will divide the side BC in the ration AB:AC i.e.,in the ratio $$7:3$$ internally .Let the bisector of $$\angle BAC$$ , meets the side BC at point D.
Therefore,D divides BC in the ratio $$7:3$$
coordinates of D are
$$\begin{array}{l} \left( { \dfrac { { 7\times 0+3\times 5 } }{ { 7+3 } } ,\dfrac { { 7\times 4+3\times 0 } }{ { 7+3 } } ,\dfrac { { 7\times \left( { -1 } \right) +3\times \left( { -6 } \right) } }{ { 7+3 } } } \right) \\ \left( { \dfrac { 3 }{ 2 } ,\dfrac { { 14 } }{ 5 } ,-\dfrac { 5 }{ 3 } } \right) \end{array}$$
Therefore, direction ratio of the bisector AD are
$$\dfrac{3}{2} - \left( { - 1} \right),\dfrac{{14}}{5} - 2,\dfrac{{ - 5}}{2} + 3\,\,\,\,i.e.\frac{5}{2},\dfrac{4}{5},\dfrac{1}{2}$$
Hence , direction cosines of the bisector AD are
$$\begin{array}{l} \dfrac { { \dfrac { 5 }{ 2 } } }{ { \sqrt { { { \left( { \dfrac { 5 }{ 2 } } \right) }^{ 2 } }+{ { \left( { \dfrac { 4 }{ 5 } } \right) }^{ 2 } }+{ { \left( { \dfrac { 1 }{ 2 } } \right) }^{ 2 } } } } } \\ \dfrac { { \dfrac { 4 }{ 5 } } }{ { \sqrt { { { \left( { \dfrac { 5 }{ 2 } } \right) }^{ 2 } }+{ { \left( { \dfrac { 4 }{ 5 } } \right) }^{ 2 } }+{ { \left( { \dfrac { 1 }{ 2 } } \right) }^{ 2 } } } } } \\ \dfrac { { \dfrac { 1 }{ 2 } } }{ { \sqrt { { { \left( { \dfrac { 5 }{ 2 } } \right) }^{ 2 } }+{ { \left( { \dfrac { 4 }{ 5 } } \right) }^{ 2 } }+{ { \left( { \dfrac { 1 }{ 2 } } \right) }^{ 2 } } } } } \\ \dfrac { { 25 } }{ { \sqrt { 714 } } } ,\dfrac { 8 }{ { \sqrt { 714 } } } ,\dfrac { 5 }{ { \sqrt { 714 } } } \end{array}$$
Question 5
1 / -0

If the points whose position vectors are $$2i+j+k, 6i-j+2k$$ and $$14i-5j+pk$$ are collinear, then the value of p is?

A
$$2$$

B
$$4$$

C
$$6$$

D
$$8$$

Solution

$$Positive\, \, vector\, \, are\, \, \, 2i+j+k,6\hat { i } +\hat { j } +2\hat { k } \\ and\, \, 4i-5j+pk\, \, are\, \, collinear\, \, then\, \, P=2 \\ if\, \, three\, \, position\, \, vectors\, \, are\, \, collinear\, \, then,\, \, its\, \, { { determinantsis } }\left( 0 \right) \\ \Rightarrow \left| \begin{matrix} 2\, \, \, \, \, \, \, 1\, \, \, \, \, \, \, 1 \\ 6\, \, \, \, -1\, \, \, \, \, \, 2 \\ 14\, \, -5\, \, \, \, P \\ \end{matrix} \right| =0 \\ \Rightarrow 2\left( { -P+10 } \right) -1\left( { 6P-28 } \right) +\left( { -30+14 } \right) =0 \\ \Rightarrow -2P+20-6P+28-16=0 \\ \Rightarrow -8P+32=0 \\ \therefore P=4\, $$
Question 6
1 / -0

The projection of a vector on three coordinate axes are $$ 6,-3, 2$$ respectively. The direction cosines of the vector are

A
$$\left(6,-3,2\right) $$

B
$$\left(\dfrac{6}{5},\dfrac{-5}{5},\dfrac{2}{5} \right) $$

C
$$\left(\dfrac{6}{7},\dfrac{-3}{7},\dfrac{2}{7}\right) $$

D
$$\left(\dfrac{-6}{7},\dfrac{-3}{7},\dfrac{2}{7}\right) $$

Solution

If $$\cos{\alpha},\cos{\beta},\cos{\gamma}$$ are the direction cosines then $$r\cos{\alpha}=6, r\cos{\beta}=-3, r\cos{\gamma}=2$$
Squaring ,$${r}^{2}{\cos}^{2}{\alpha}=36,{r}^{2}{\cos}^{2}{\beta}=9,{r}^{2}{\cos}^{2}{\gamma}=4$$
$$ \Rightarrow {r}^{2}{\cos}^{2}{\alpha}+{r}^{2}{\cos}^{2}{\beta}+{r}^{2}{\cos}^{2}{\gamma}=\left(36+9+4\right)=49$$
$$\Rightarrow {r}^{2}\left({\cos}^{2}{\alpha}+{\cos}^{2}{\beta}+{\cos}^{2}{\gamma}\right)=49$$
$$\Rightarrow {r}^{2}=49 , r=7$$
$$\therefore $$ the direction cosines are $$\dfrac{6}{7},\dfrac{-3}{7},\dfrac{2}{7}$$
Question 7
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The vector form of the equation of the line passing through points $$(3,4, 7)$$ and $$(5,1,6)$$ is-

A
$$\vec { r } =(3\hat { i } +4\hat { j } -7\hat { k } )+\lambda (2\hat { i } -3\hat { j } +13\hat { k } )$$

B
$$\vec { r } =(3\hat { i } +4\hat { j } -7\hat { k } )+\lambda (8\hat { i } +5\hat { j } -\hat { k } )$$

C
$$\vec { r } =(3\hat { i } +4\hat { j } +7\hat { k } )+\lambda (2\hat { i } -3\hat { j } -\hat { k } )$$

D
$$\vec { r } =(3\hat { i } +4\hat { j } -7\hat { k } )+\lambda (2\hat { i } -3\hat { j } -13\hat { k } )$$

Solution

We have,

Given points are $$\left( 3,4,7 \right)\,\,and\,\,\left( 5,1,6 \right)$$.

We know that,

Vector equation of a line passing through the two points with position vectors $$\overrightarrow{a}$$ and $$\overrightarrow{b}$$ is

$$\overrightarrow{r}=\overrightarrow{a}+\lambda \left( \overrightarrow{b}-\overrightarrow{a} \right)$$

Let the given points are $$A\left( 3,4,7 \right)\,\,and\,\,\left( 5,1,6 \right)$$

Now,

Equation of vector,

$$ \overrightarrow{r}=\left( 3\overrightarrow{i}+4\overrightarrow{j}+7\overrightarrow{k} \right)+\lambda \left( 5\overrightarrow{i}+\overrightarrow{j}+6\overrightarrow{k}-3\overrightarrow{i}-4\overrightarrow{j}-7\overrightarrow{k} \right) $$

$$ \overrightarrow{r}=\left( 3\overrightarrow{i}+4\overrightarrow{j}+7\overrightarrow{k} \right)+\lambda \left( 2\overrightarrow{i}-3\overrightarrow{j}-\overrightarrow{k} \right) $$

Hence, this is the answer.
Question 8
1 / -0

If two lines $$x=2y+3, z=y+1$$ and $$x=4y+1, z=3y+2$$ makes an angle $$\theta$$ with each other, then $$\cos\theta$$ equals

A
$$\dfrac{5}{\sqrt{39}}$$

B
$$\dfrac{3}{\sqrt{13}}$$

C
$$\dfrac{3}{\sqrt{39}}$$

D
$$\dfrac{6}{\sqrt{39}}$$
Question 9
1 / -0

The line perpendicular to the plane $$2x-y+5z=4$$ passing through the point $$(-1,0,1)$$ is ?

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

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

C
$$\cfrac{x+1}{2}=-y=\cfrac{z-1}{5}$$

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

Solution

$$\begin{array}{l} Po{ { int } }\, \, A\left( { -1,0,1 } \right) \, and\, \, \, B\left( { 2,-1,5 } \right) \\ So,\, \, equation\, \, of\, \, line. \\ \frac { { x+1 } }{ 2 } =-y=\frac { { z-1 } }{ 5 } \\ Option\, \, \, C\, \, is\, \, correct\, \, answer. \end{array}$$
Question 10
1 / -0

If a straight line makes an angle $${ cos }^{ -1 }\left( \frac { 1 }{ \sqrt { 3 } } \right) $$ with each of the positive $$x, y$$ and $$z$$-axis, a vector parallel to that line is

A
$$\overset { - }{ i } $$

B
$$\overline { i } +\overline { j } $$

C
$$\overline { j } +\overline { k } $$

D
$$\overline { i } +\overline { j } +\overline { k } $$

Solution

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Re-Attempt Weekly Quiz Competition

Three Dimensional Geometry Test - 46

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Three Dimensional Geometry Test - 46

Score

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SHARING IS CARING

Question 1

atleast one $$x$$ is $$\left(0,1\right)$$

atleast one $$x$$ is $$\left(-1,0\right)$$

atleast one $$x$$ is $$\left(1,2\right)$$

none of these

Solution

Question 2

$$0$$

$$\alpha\overrightarrow{a}$$

$$\beta\overrightarrow{b} $$

$$\left(\alpha+\beta\right)\overrightarrow{c}$$

Solution

Question 3

$$ \frac{2}{5} $$

$$ \frac{5}{3} $$

$$ \frac{2}{3} $$

$$ \frac{3}{2} $$

Solution

Question 4

$$\left(\dfrac{25}{\sqrt{714}}, \dfrac{-8}{\sqrt{714}}, \dfrac{-5}{\sqrt{714}}\right)$$

$$\left(\dfrac{5}{\sqrt{74}}, \dfrac{6}{\sqrt{74}}, \dfrac{8}{\sqrt{74}}\right)$$

$$\left(\dfrac{25}{\sqrt{714}}, \dfrac{8}{\sqrt{714}}, \dfrac{5}{\sqrt{714}}\right)$$

$$\left(\dfrac{-5}{\sqrt{74}}, \dfrac{6}{\sqrt{74}}, \dfrac{-8}{\sqrt{74}}\right)$$

Solution

Question 5

$$2$$

$$4$$

$$6$$

$$8$$

Solution

Question 6

$$\left(6,-3,2\right) $$

$$\left(\dfrac{6}{5},\dfrac{-5}{5},\dfrac{2}{5} \right) $$

$$\left(\dfrac{6}{7},\dfrac{-3}{7},\dfrac{2}{7}\right) $$

$$\left(\dfrac{-6}{7},\dfrac{-3}{7},\dfrac{2}{7}\right) $$

Solution

Question 7

$$\vec { r } =(3\hat { i } +4\hat { j } -7\hat { k } )+\lambda (2\hat { i } -3\hat { j } +13\hat { k } )$$

$$\vec { r } =(3\hat { i } +4\hat { j } -7\hat { k } )+\lambda (8\hat { i } +5\hat { j } -\hat { k } )$$

$$\vec { r } =(3\hat { i } +4\hat { j } +7\hat { k } )+\lambda (2\hat { i } -3\hat { j } -\hat { k } )$$

$$\vec { r } =(3\hat { i } +4\hat { j } -7\hat { k } )+\lambda (2\hat { i } -3\hat { j } -13\hat { k } )$$

Solution

Question 8

$$\dfrac{5}{\sqrt{39}}$$

$$\dfrac{3}{\sqrt{13}}$$

$$\dfrac{3}{\sqrt{39}}$$

$$\dfrac{6}{\sqrt{39}}$$

Question 9

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

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

$$\cfrac{x+1}{2}=-y=\cfrac{z-1}{5}$$

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

Solution

Question 10

$$\overset { - }{ i } $$

$$\overline { i } +\overline { j } $$

$$\overline { j } +\overline { k } $$

$$\overline { i } +\overline { j } +\overline { k } $$

Solution

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