Tag: maths

Questions Related to maths

The point collinear with $(4, 2, 0)$ and $(6, 4, 6)$ among the following is

direction cosines and direction ratios three dimensional geometry maths

  1. $(0,4,6)$

  2. $(8,6,8)$

  3. $(1, -4, -6)$

  4. None of these

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Correct Option: D
Explanation:
Cartesian form  of a line passing through $(x _1,y _1,z _1)$  and  $(x _2,y _2,z _2) $ is:
$\dfrac{x-x _1}{x _2-x _1}=\dfrac{y-y _1}{y _2-y _1}=\dfrac{z-z _1}{z _2-z _1}=\lambda$

Cartesian form  of the line passing through (4,2,0) and (6,4,6) is:
$\dfrac{x-4}{6-4}=\dfrac{y-2}{4-2}=\dfrac{z-0}{6-0}=\lambda$

$\dfrac{x-4}{2}=\dfrac{y-2}{2}=\dfrac{z-0}{6}=\lambda$

Therefore general point on the line is $(2\lambda+4,2\lambda+2,6\lambda)$
Now if we compare this coordinate with given options,we get different values of $\lambda$.

that means none of the points given in options lie on the line(Non-Colinear).

Therefore, (D) option is correct.

If the points $(0, 1, -2), (3$, $\lambda$,$ 1)$ and ($\mu$, $7, 4$) are collinear, the point on the same line is

direction cosines and direction ratios three dimensional geometry maths

  1. $(5, 6, 3)$

  2. $(1, -1, -2)$

  3. $(-5, -6, -3)$

  4. $(0, 0, 0)$

Show answer
Correct Option: A
Explanation:
The direction vector using the first two points we get $3i+(\lambda-1)j+3k$. 
Using the first and third point $\mu i+6j+6k$. 
Since they are up to multiplication by a constant we get $\lambda=4$ and $\mu=6$.
 Hence direction vector is $i+j+k$. 
Any point on the line is given by $(0,1,-2)+t(1,1,1)$. Plugging $t=5$ we get the point $(5,6,3)$.   

Given $A(1,-1,0)$; $B(3,1,2)$;$C(2,-2,4)$ and $D(-1,1,-1)$ which of the following points neither lie on $AB$ nor on $CD$

direction cosines and direction ratios three dimensional geometry maths

  1. $(2,2,4)$

  2. $(2,-2,4)$

  3. $(2,0,1)$

  4. $(0,-2,-1)$

Show answer
Correct Option: A
Explanation:

Given:- $A(1,-1,0)$; $B(3,1,2)$;$C(2,-2,4)$ and $D(-1,1,-1)$ 

The equation of line $AB$ is given by $r=i-j+t(2i+2j+2k)$ or $\dfrac{x-1}{2} = \dfrac{y+1}{2} = \dfrac{z}{2} $ 
The points $(2,0,1)$ and $(0,-2,1)$ lies on $AB$. 
The equation of line $CD$ is given by $r=2i-2j+4k+t(3i-3j+5k)$ or $ \dfrac{x-2}{3} = \dfrac{y+2}{-3} = \dfrac{z-4}{5}$
$(2,-2,4)$ lie on $CD$. 
$(2,2,4)$ does not lie on any of the lines AB or CD.
Hence, option A is correct.

If the points $a(1, 2, -1), B(2, 6, 2)$ and $c(\lambda, -2, -4)$ are collinear then $\lambda$ is

direction cosines and direction ratios three dimensional geometry maths

  1. $0$

  2. $2$

  3. $-2$

  4. $1$

Show answer
Correct Option: A
Explanation:

D.R of AB are $ 2 -1, 6-2,2-(-1) i.e. 1,4,3$

D.R. of AC are $ λ.−1,−2−2,−4−(−1) $
$i.e., λ−1,−4,−3 $
Since A, B, C are collinear $\therefore AB||BC $
$\therefore \dfrac {\lambda-1}{1} =\dfrac{-4}{4}=\dfrac{-3}{3}$
$\Longrightarrow \lambda-1=-1$
$\therefore \lambda=0$

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

direction cosines and direction ratios three dimensional geometry maths

  1. -1

  2. 1

  3. 2

  4. 0

Show answer
Correct Option: B
Explanation:
If the area of triangle is zero, then the points are collinear.
Points are collinear.
Point are $(p, o)(o, q)(i,q)$

$\Delta =\dfrac {1}{2}[p(q-1)-0(1-0)+1(0-q)]$
$\Rightarrow \dfrac{1}{2}[p(q-1)-q]$
$\Rightarrow \dfrac{1}{2}[p(q-1)-q]$

$\Rightarrow \dfrac{1}{2}[pq-(p+q)]=o$
$\Rightarrow pq=p+q\Rightarrow \dfrac {p+q}{pq}=1$

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

Given $A(1,-1,0)$; $B(3,1,2)$; $C(2,-2,4)$ and $D(-1,1,-1)$ which of the following points neither lie on $AB$ nor on $CD$?

direction cosines and direction ratios three dimensional geometry maths

  1. $(2,2,4)$

  2. $(2,-2,4)$

  3. $(2,0,1)$

  4. $(0,-2,-1)$

Show answer
Correct Option: A
Explanation:

$A(1,-1,0) , B(3,1,2), C(2,-2,4), D(-1,1,-1)\ \vec { AB } = <3-1, 1-(-1), 2-0>$ 

       $= <2,2,2>$ 
$\therefore \quad Equation\quad of\quad line\quad AB:\ \dfrac { x-1 }{ 2 } =\dfrac { y-(-1) }{ 2 } =\dfrac { z-0 }{ 2 } \quad \Longrightarrow \quad \dfrac { x-1 }{ 2 } =\dfrac { y+1 }{ 2 } =\dfrac { z-0 }{ 2 }$ 
$\vec { CD } = <-1-2, 1-(-2), -1-4>$ 
        $= <-3,3,-5>$ 
$\therefore \quad Equation\quad of\quad line\quad CD:\ \dfrac { x-2 }{ -3 } =\dfrac { y-(-2) }{ 3 } =\dfrac { z-4 }{ -5 } \quad \Longrightarrow \quad \dfrac { x-2 }{ -3 } =\dfrac { y+2 }{ 3 } =\dfrac { z-4 }{ -5 }$ 
By putting the values of points given in the choices in the equation of lines $AB$ and $CD$, $(2,2,4)$ is the point which neither lie on $AB$ nor on $CD$.

If the points $A(1,2,-1)$, $B(2,6,2)$ and $\displaystyle C\left ( \lambda,-2,-4 \right )$ are collinear, then $\displaystyle \lambda $ is

direction cosines and direction ratios three dimensional geometry maths

  1. $0$

  2. $2$

  3. $-2$

  4. $1$

Show answer
Correct Option: A
Explanation:

D.R. of AB are $2−1,6−2,2−(−1)i.e.,1,4,3.$

D.R. of AC are $λ.−1,−2−2,−4−(−1) $
$i.e., λ−1,−4,−3 $
Since A, B, C are colinear, 
$ \therefore AB||BC$
$\therefore \dfrac{λ−1}{1}=\dfrac{−4}{4}=\dfrac{−3}{3}⇒λ−1=−1⇒λ=0$

The position vectors of three points are $2\vec{a}-\vec{b}+3\vec{c}$, $\vec{a}-2\vec{b}+\lambda \vec{c}$ and $\mu \vec{a}-5\vec{b}$ where $\vec{a}, \vec{b}, \vec{c}$ are non coplanar vectors, then the points are collinear when

direction cosines and direction ratios three dimensional geometry maths

  1. $\displaystyle \lambda =-2, \mu =\dfrac{9}{4}$

  2. $\displaystyle \lambda =-\dfrac{9}{4}, \mu =2$

  3. $\displaystyle \lambda =\dfrac{9}{4}, \mu =-2$

  4. None of these

Show answer
Correct Option: C
Explanation:

When points $x, y, z$ are collinear, we have $\alpha x + \beta y = (\alpha + \beta)z$
Similarly, $x(2\vec{a} - \vec{b} + 3\vec{c}) + y(\vec{a} - 2\vec{b} + \lambda \vec{c}) = (x + y)(\mu \vec{a} - 5\vec{b})$
$\Rightarrow$ comparing the coefficients of $\vec{a} \rightarrow 2x + y = x\mu + y\mu $
$\vec{b} \rightarrow - x - 2y = - 5x - 5y$
$\vec{c} \rightarrow 3x + \lambda y = 0$
$\Rightarrow 4x = -3y$ and so $\lambda = \dfrac{9}{4}$
Also, $\mu = -2$

$\bar a,\bar b,\bar c$ are three non-zero vectors such that any two of them are non-collinear. If  $\bar a+\bar b$ is collinear with  $\bar c$ and  $\bar b+\bar c$ is collinear with $\bar a$, then what is their sum?

direction cosines and direction ratios three dimensional geometry maths

  1. $-1$

  2. $0$

  3. $1$

  4. $2$

Show answer
Correct Option: B
Explanation:

We have

$\bar a+\bar b =t\bar c$ ----$(1)$
$\bar b+\bar c =s\bar a$ ----$(2)$
From $(1)$ and $(2)$
$\bar a+\bar b=t(s\bar a-\bar b)$
Since no two of them are collinear, comparing coeffficients gives
$st=1$ and $t=-1$
$\Rightarrow s=-1$ and $t=-1$
From $(1)$
$\therefore \bar a+\bar b+\bar c=0$
Hence, option $B$.

The line passes through the points $\left ( 5,1,a \right )$ & $\left ( 3,b,1 \right )$ crosses the $yz$ plane at the point $\displaystyle \left ( 0,\frac{17}{2},-\frac{13}{2} \right )$ ,then

direction cosines and direction ratios three dimensional geometry maths

  1. $a= 4, b= 6$

  2. $a= 6, b= 4$

  3. $a= 8, b= 2$

  4. $a= 2, b= 8$

Show answer
Correct Option: B
Explanation:

Equation of the line through the points $( 5,1,a  ) $ & $( 3,b,1)$ is
$\displaystyle \frac{x-5}{3-5}=\frac{y-1}{b-1}=\frac{z-a}{1-a}=\lambda $
Now it passes through $\displaystyle \left ( 0,\frac{17}{2},\frac{-13}2{} \right )$
$\displaystyle \therefore \frac{0-5}{-2}=\frac{17/2-1}{b-1}=\frac{-13/2-a}{1-a}=\lambda  :$

$ \Rightarrow \lambda =\dfrac{5}{2}$
$\displaystyle \therefore \frac{17/2-1}{b-1}=\frac{5}{2} $
$\Rightarrow b=4$
and $\displaystyle \frac{-13/2-a}{1-a}=\frac{5}{2} $
$\Rightarrow a=6$