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Questions Related to maths

Multiple choice direction cosines and direction ratios three dimensional geometry maths

If the three points with position vectors $\displaystyle \bar{a}-2\bar{b}+3\bar{c}, \ 2\bar{a}+\lambda \bar{b}-4\bar{c}, \ -7\bar{b}+10\bar{c} $ are collinear, then $\displaystyle \lambda= $

  1. <font color="#888888">$1$</font>

  2. <span class="MathJax_Preview"><span class="MJXp-math"><span class="MJXp-mn">2
  3. $3$

  4. none of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The given vectors are collinear, so $l(\bar{a} - 2\bar{b} + 3\bar{c}) + k(2\bar{a} + \lambda\bar{b} - 4\bar{c}) = (l + k)(-7\bar{b} + 10\bar{c})$
Comparing the coefficients of $\bar{a} \rightarrow l + 2k = 0 $
$\bar{b} \rightarrow -2l + \lambda k = -7l -7k$
$\bar{c} \rightarrow 3l - 4k = 10l + 10k$
$\Rightarrow l = -2k$ and so $\lambda = 3$

Multiple choice direction cosines and direction ratios three dimensional geometry maths

The vectors $2\hat i + 3\hat j, \ 5\hat i + 6\hat j$ and $8\hat i + \lambda \hat j$ have their initial points at $(1,1)$. The value of $\lambda$ so that the vectors terminate on one straight line is

  1. 9

  2. 6

  3. 3

  4. 0
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Vectors starting from the same point (1,1) are collinear if their components are proportional. The vectors are v1 = (2, 3), v2 = (5, 6), and v3 = (8, lambda). The vector v2 - v1 = (3, 3). The vector v3 - v2 = (3, lambda - 6). For these to be collinear, the slopes must be equal, so (lambda - 6) / 3 = 3 / 3, which gives lambda - 6 = 3, so lambda = 9.

Multiple choice direction cosines and direction ratios three dimensional geometry maths

For what value of $m$, the points $(3,5)$, $(m,6)$ and $\begin{pmatrix} \dfrac { 1 }{ 2 },\dfrac {15 }{ 2 } \end{pmatrix}$ are collinear?

  1. $9$
  2. $5$
  3. $3$
  4. $2$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

As the points are collinear, the slope of the line joining any two points, should be same as the slope of the line joining two other points. 
Slope of the line passing through points $\left( { x } _{ 1 },{ y } _{ 1 } \right) $ and $\left( { x } _{ 2 },{ y } _{ 2 } \right)$ $ $=$ $ $\dfrac { { y } _{ 2 }-{ y } _{ 1 } }{ { x } _{ 2 }-x _{ 1 } } $
So, slope of the line joining $ (3,5) , (m,6) = $ Slope of the line joining $ (3,5) $ and $\left  (\dfrac {1}{2}, \dfrac {15}{2}\right ) $ 

Therefore, $ \dfrac { 6 - 5 }{ m - 3 } = \dfrac { \frac {15}{2} - 5 }{ \frac {1}{2} - 3 } $
$\Rightarrow  \dfrac { 1 }{ m - 3 } = -1 $

$\Rightarrow  m - 3 = -1 $

$\Rightarrow  m = 2 $

Multiple choice direction cosines and direction ratios three dimensional geometry maths

If the points $(p,0)$, $(0,q)$ and $(1,1)$ are collinear, then $\dfrac { 1 }{ p }+\dfrac { 1 }{ q }$ is equal to:

  1. $-1$
  2. $1$
  3. $2$
  4. $0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

As the points are collinear, the slope of the line joining

any two points, should be same as the slope of the line joining two other

points.

Slope of the line passing through points $\left( { x } _{ 1 },{ y } _{ 1 }

\right) $ and $\left( { x } _{ 2 },{ y } _{ 2 } \right)$ $ = $ $\dfrac { { y

} _{ 2 }-{ y } _{ 1 } }{ { x } _{ 2 }-x _{ 1 } } $

So, slope of the line joining $ (p,0) , (0,q) = $ Slope of the line joining

$ (0,q) $ and $ (1,1) $

$ \dfrac { q - 0 }{ 0 - p } = \dfrac { 1 - q }{ 1 - 0 } $

$ - \dfrac { q }{ p } = 1 - q $

Dividing both sides by $q$,
$ - \dfrac { 1 }{ p } =  \dfrac { 1 }{ q } - 1 $

$ => \dfrac { 1 }{ p } +  \dfrac { 1 }{ q } = 1 $

Multiple choice direction cosines and direction ratios three dimensional geometry maths

Determine if the points $(1,5)$ $(2,3)$ and $(-2,-11)$ are collinear.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The given points are $A(1,5)$, $B(2,3)$ and $C(-2,-11)$.


Let us calculate the distance : $AB$, $BC$ and $CA$ by using distance formula.

$AB =\sqrt { (2-1)^{ 2 }+(3-5)^{ 2 } } =\sqrt { (1)^{ 2 }+(-2)^{ 2 } } $

$=\sqrt {1+4} = \sqrt{ 5 }$ units

$BC =\sqrt { (-2-2)^{ 2 }+(-11-3)^{ 2 } }=\sqrt { (-4)^{ 2 }+(-14)^{ 2 } }$

$=\sqrt {16+196} =\sqrt {212} = 2\sqrt{53}$ units

$CA =\sqrt { (-2-1)^{ 2 }+(-11-5)^{ 2 } }$

$=\sqrt { (-3)^{ 2 }+(-16)^{ 2 } } =\sqrt {9+256} = \sqrt {265 }$ 

$=\sqrt {5}\times\sqrt {53}$ units

From the above we see that : $AB+BC\neq CA$

Hence, the above stated points $A(1,5)$, $B(2,3)$ and $C(-2,-11)$ are not collinear.

Multiple choice direction cosines and direction ratios three dimensional geometry maths

In each of the following find the value of $k$, for which the points are collinear.
(i) $(7,-2)$, $(5,1)$, $(3,k)$
(ii) $(8,1)$, $(k,-4)$, $(2,-5)$

  1. (i) $k = 4$
  2. (i) $k = 5$
  3. (ii) $k = 3$
  4. (ii) $k = 2$
Reveal answer Fill a bubble to check yourself
A,C Correct answer
Explanation

Since the given points are collinear, they do not form a triangle, which means area of the triangle is Zero.

Area of a triangle with vertices $({ x } _{ 1 },{ y } _{ 1 })$ ; $({ x } _{ 2 },{ y

} _{ 2 })$  and $({ x } _{ 3 },{ y } _{ 3 })$  is $ \left| \dfrac { {

x } _{ 1 }({ y } _{ 2 }-{ y } _{ 3 })+{ x } _{ 2 }({ y } _{ 3 }-{ y } _{ 1 })+{ x } _{

3 }({ y } _{ 1 }-{ y } _{ 2 }) }{ 2 }  \right| $


1) Substituting the points $({ x } _{ 1 },{ y } _{ 1 }) = (7,-2) $ ; $({ x

} _{ 2 },{ y } _{ 2 }) = (5,1) $  and $({ x } _{ 3 },{ y } _{ 3 }) = (3,k)$

In the area formula, we get

$ \left| \dfrac { 7(1-k) + 5(k+2) + 3(-2-1) }{ 2 }  \right|  =

0 $

$ \left| \dfrac { 7 -7k + 5k + 10 - 9 }{ 2 }  \right|  =

0 $

$ \left| \dfrac { 8 -2k }{ 2 }  \right|  =

0 $

$ \Rightarrow  8 - 2k = 0 $

$ \Rightarrow  k = 4 $

2) Substituting the points $({ x } _{ 1 },{ y } _{ 1 }) = (8,1) $ ; $({ x

} _{ 2 },{ y } _{ 2 }) = (k,-4) $  and $({ x } _{ 3 },{ y } _{ 3 }) = (2,-5)$ in the area formula, we get


$ \left| \dfrac { 8(-4+5) + k(-5-1) + 2(1+4) }{ 2 }  \right|  =

0 $

$ \left| \dfrac { 8 -6k +10 }{ 2 }  \right|  =

0 $

$ \left| \dfrac { 18 -6k }{ 2 }  \right|  =

0 $

$ \Rightarrow  18 - 6k = 0 $

$ \Rightarrow  k = 3 $

Multiple choice direction cosines and direction ratios three dimensional geometry maths

Are the points (1, 1), (2, 3) and (8, 11) collinear ?

  1. collinear

  2. Non collinear

  3. coplaner

  4. None of above
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Area of triangle formed by these vertices is 
$\displaystyle \Delta =\frac { 1 }{ 2 } \begin{vmatrix} 1 & 1 & 1 \ 2 & 3 & 1 \ 8 & 11 & 1 \end{vmatrix}$
Applying ${ R } _{ 2 }\rightarrow { R } _{ 2 }-{ R } _{ 1 },{ R } _{ 3 }\rightarrow { R } _{ 3 }-{ R } _{ 1 }$
$\displaystyle \Delta =\frac { 1 }{ 2 } \begin{vmatrix} 1 & 1 & 1 \ 1 & 2 & 0 \ 7 & 10 & 0 \end{vmatrix}=\frac { 1 }{ 2 } \left( 10-14 \right) =2$
Hence points are non collinear 

Multiple choice direction cosines and direction ratios three dimensional geometry maths

If $\vec{a},\vec{b},\vec{c}$ are the position vectors of points lie on a line, then $\vec{a}\times \vec{b}+\vec{b}\times \vec{c}+\vec{c}\times \vec{a}=$

  1. $0$
  2. $ \vec{b}$
  3. $1$
  4. $\vec{a}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If points with position vectors a, b, c are collinear, then (b-a) is parallel to (c-b). This implies (b-a) x (c-b) = 0. Expanding this cross product gives b x c - b x b - a x c + a x b = 0. Since b x b = 0, we get b x c + a x b - a x c = 0, which rearranges to a x b + b x c + c x a = 0.

Multiple choice direction cosines and direction ratios three dimensional geometry maths

Assertion ($A$): The points with position vectors $\overline{a},\overline{b},\overline{c}$ are collinear if $2\overline{a}-7\overline{b}+5\overline{c}=0$.
Reason ($R$): The points with position vectors $\overline{a},\overline{b},\overline{c}$ are collinear if $l\overline{a}+m\overline{b}+n\overline{c}=\overline{0}$.

  1. Both $A$ and $R$ are true and $R$ is correct reason of $A$
  2. Both $A$ and $R$ are true and $R$ is not correct reason of $A$
  3. $A$ is true $R$ is false
  4. $A$ is false $R$ is true
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\overrightarrow { a } ,\overrightarrow { b } ,\overrightarrow { c } $ are collinear

$\Rightarrow \ni l,m,n$ all zeros such that 
$l\overrightarrow { a } +m\overrightarrow { b } +n\overrightarrow { c } =0$ if $2\overrightarrow { a } -7\overrightarrow { b } +5\overrightarrow { c } =0$
then $\overrightarrow { a } ,\overrightarrow { b } ,\overrightarrow { c } $ are collinear since $2-7+5=0$

Multiple choice direction cosines and direction ratios three dimensional geometry maths

The points with position vectors $\vec{a}+\vec{b},\vec{a}-\vec{b}$ and $\vec{a}+\lambda\vec{b}$ are collinear for

  1. Only integrals values of $\lambda$
  2. No value of $\lambda$
  3. All real values of $\lambda$
  4. Only rational values of $\lambda$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Three points with position vectors p1, p2, p3 are collinear if (p2-p1) is a multiple of (p3-p2). Here, p2-p1 = -2b and p3-p2 = (lambda-1)b. Since both vectors are multiples of b, they are parallel for any real value of lambda.