Tag: three dimensional geometry

Quizzes (19) Questions (122)

Questions Related to three dimensional geometry

Multiple choice direction cosines and direction ratios three dimensional geometry maths

If $\vec { a } ,\vec { b } ,\vec { c } $ are three non-zero vectors, no two of which are collinear and the vector $\vec { a } +\vec { b } $ is collinear with $\vec { c }, \vec { b } +\vec { c } $ is collinear with $\vec {a},$ then $\vec { a } +\vec { b } +\vec { c }$ is equal to -

  1. $\vec {a}$
  2. $\vec {b}$
  3. $\vec {c}$
  4. $none\ of\ these$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
$\bar a+\bar b=K _{1}\bar c $

$\bar b+\bar c=K _{2}\bar a $

$\bar a- \bar c=K _{1}c-K _{2}\bar a$

$(k _{2}+1) \bar a-\bar c(1+k _{1})$=0

$k _{2}=-1 $ and $k _{1}=-1 $

$\bar{a}+\bar{b}+\bar{c}=0$
Multiple choice direction cosines and direction ratios three dimensional geometry maths

If the points with position vectors $60\hat{i}+3\hat{j}, 40\hat{i}-8\hat{j}$ and $a\hat{i}-52j$ are collinear, then $a=?$

  1. $-40$
  2. $-20$
  3. $20$
  4. $40$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Suppose, position vector $A=60\widehat i+3\widehat j$

position vector $B=40\widehat i-8\widehat j$

position vector $C=a\widehat i-52\widehat j$

Now, find vector AB and BC

$AB= -20\widehat i-11\widehat j$

$BC= (a-40)\widehat i-44\widehat j$

To be collinear,  angle between the vector AB and BC made by the given position vectors should be 0 or 180 degree.

That’s why the cross product of  the vectors should be zero

$ABXBC=(-20\widehat i-11\widehat j)X(a-40)\widehat i-44\widehat j$

$0\widehat i+0\widehat j+(880+11(a-40))=0$

$a-40= -80$

$a=-40$

Therefore, a should be $-40$ to be the given positions vectors collinear.


Multiple choice direction cosines and direction ratios three dimensional geometry maths

 Let $\overrightarrow{b}$ and  $\overrightarrow{c}$ be non collinear vectors.If $\overrightarrow{a}$ is a vector such that $\overrightarrow{a}.\left(\overrightarrow{b}+\overrightarrow{c}\right)=4$ and $\overrightarrow{a}\times\left(\overrightarrow{b}\times \overrightarrow{c}\right)=\left({x}^{2}-2x+6\right)\overrightarrow{b}+\sin{y} .\overrightarrow{c}$ then $\left(x,y\right)$ lies on the line

  1. $x+y=0 $
  2. $x-y=0$
  3. $x=1$
  4. $y=\dfrac{\pi}{2}$
Reveal answer Fill a bubble to check yourself
C,D Correct answer
Explanation

$\overrightarrow{a}\times \left(\overrightarrow{b}\times \overrightarrow{c}\right)=\left(\overrightarrow{a}.\overrightarrow{c}\right).\overrightarrow{b}-\left(\overrightarrow{a}.\overrightarrow{b}\right).\overrightarrow{c}$
$\therefore \overrightarrow{a}.\overrightarrow{c}={x}^{2}-2x+6=-\sin y$
$\overrightarrow{a}.\left(\overrightarrow{b}+\overrightarrow{c}\right)=4 \Rightarrow -\sin y+{x}^{2}-2x+6=4$
$\Rightarrow {x}^{2}-2x+2=\sin y$
$\Rightarrow {\left(x-1\right)}^{2}+1=\sin y$
Left side $\ge 1$, right side $\le 1$
$\therefore $ they are equal if 
${\left(x-1\right)}^{2}+1=\sin y=1$
$\therefore y=\dfrac{\pi}{2},x=1$

Multiple choice direction cosines and direction ratios three dimensional geometry maths

Three points whose position vectors are $x\bar{i}+y\bar{j}+z\bar{k}$, $\bar{i}+2\bar{j}$ and $-\bar{i}-\bar{j}$ are collinear, then relation between $x, y, z$ is?

  1. $x-2y=1, z=0$
  2. $z+y=1, z=0$
  3. $x-y=1, z=0$

  4. None of these
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For three points to be collinear, the vectors connecting them must be proportional. The vectors are P1(x, y, z), P2(1, 2, 0), and P3(-1, -1, 0). The vector P2P3 is (-2, -3, 0). The vector P1P2 is (1-x, 2-y, -z). For these to be parallel, the ratios of components must be equal, implying z=0 and a specific linear relationship between x and y that is not listed in A, B, or C.

Multiple choice direction cosines and direction ratios three dimensional geometry maths

If the points $(\alpha, - 1), (2, 1)$ and $(4, 5)$ are collinear, then find $\alpha $ by vector method.

  1. $4$
  2. $1$
  3. $8$

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
If there points are collinear then vectors from one to another will have scalar triple produced $0$.Point $\left(\alpha,-1\right), \left(2,1\right), \left(4,5\right)$
$\left( 2-\alpha  \right) \hat { i } +2\hat { j } -\bar { A }$
$2\hat { i } +4\hat { j } -\bar { B }$
$\left( 4-\alpha  \right) \hat { i } +6\hat { j } -\bar { C }$
$ \bar { A } .\left( \bar { B } \times \bar { C }  \right) =0$
$\left( \left( 2-\alpha  \right) \hat { i } +2\hat { j }  \right) \left( 2\hat { i } +4\hat { j }  \right) \times \left( \left( 4-\alpha  \right) \hat { i } +6\hat { j }  \right) \\ \left( \left( 2-\alpha  \right) \hat { i } +2\hat { j }  \right) .\left[ 12\hat { k } -16\hat { k } +4\alpha \hat { k }  \right] =0$
$4\alpha =4$
 $\alpha =1$
Also the direction vector will be proportion
$\left( 2-\alpha,2 \right)=\lambda\left( 4-2.5-1\right)$
$\left( 2-\alpha,2 \right)=\lambda\left( 2,4\right)$
$\lambda=\dfrac{1}{2}$ as $2=4\lambda$
$2-\alpha=1$
$\therefore \alpha=1$

Multiple choice direction cosines and direction ratios three dimensional geometry maths

If the points $\bar a + \bar b,\bar a - \bar b,\bar a + k\bar b$ are collinear, then  

  1. $k$ has only one real value
  2. $k$ has two real value
  3. $k$ has no real values
  4. $k$ has infinite number of real values
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
As the $3$ point should be collinear area of triangle termed by then should be zero

considering $2$ direction to be $(\bar{a}+\bar{b})-(\bar{a}-\bar{b})=2\bar{b}$
& $(\bar{a}+\bar{b})-(\bar{a}+k\bar{b})=(1-k)\bar{b}$

$(2\bar{b})\times (1-k)\bar{b}=0$

this will be for any value of $k$ as cross produced of $2$ linear vector $=0$


Multiple choice direction cosines and direction ratios three dimensional geometry maths

If $A = (1,2,3) , B  = (2,10,1), Q$ are collinear points and $Q _{x}=-1$ then $Q _{z}$ is

  1. $-3$
  2. $7$
  3. $-14$
  4. $-7$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Since A(1, 2, 3), B(2, 10, 1), and Q(Qx, Qy, Qz) are collinear, the vector AB = (1, 8, -2) must be proportional to AQ = (Qx-1, Qy-2, Qz-3). Given Qx = -1, the x-component of AQ is -2. Since -2 is -2 times the x-component of AB, the other components must follow the same ratio: Qy-2 = -2(8) = -16 (Qy = -14) and Qz-3 = -2(-2) = 4 (Qz = 7).

Multiple choice direction cosines and direction ratios three dimensional geometry maths

If points $\hat i + \hat j, \hat i - \hat j$ and $p \hat i + q \hat j + r \hat k$ are collinear, then

  1. $p = 1$
  2. $r = 0$
  3. $q \in R$
  4. $q \neq 1$
Reveal answer Fill a bubble to check yourself
A,B,D Correct answer
Explanation

Points $A(\hat i + \hat j), B (\hat i - \hat j)$ and $C(p \hat i + q \hat j + r \hat k)$ are collinear
Now $\vec{AB} = - 2 \hat j$ and $\vec{BC} = (p -1) \hat i + (q - 1) \hat j + r k$
Vectors $\vec{AB}$ and $\vec{BC}$ must be collinear
$\Rightarrow p = 1,  r= 0 $ and $q \neq 1$

Multiple choice direction cosines and direction ratios three dimensional geometry maths

If  $\bar { a }, \bar { b }, \bar { c }$ are non-coplaner vector , then the vectors $2\bar { a }- 4\bar { b }+ 4\bar { c }, \bar { a }- 2\bar { b }+ 4\bar { c }$ and $-\bar { a }+ 2\bar { b }+ 4\bar { c }$ are parellel.

  1. True

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

Three vectors are parallel if they are scalar multiples of each other. Here, the vectors are v1 = 2a - 4b + 4c, v2 = a - 2b + 4c, and v3 = -a + 2b + 4c. Since the coefficients of a and b are not proportional to the constant c across all three vectors, they cannot be parallel.

Multiple choice direction cosines and direction ratios three dimensional geometry maths

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

  1. $(12, 9, 2)$
  2. $(1, -1, -2)$
  3. $(5, -3, 4)$
  4. $(0, 0, 0)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
We know condition of collinearity,
$\dfrac{x-x _1}{x _2-x _1}=\dfrac{y-y _1}{y _2-y _1}=\dfrac{z-z _1}{z _2-z _1}$


We have points $(0,1,-2),(3,\lambda,-1) and (\mu,-3,-4)$
$\dfrac{x-0}{3-0}=\dfrac{y-1}{\lambda-1}=\dfrac{z+2}{-1+2}.........(1)$

Since $(\mu,-3,-4)$ is collinear , so it satisfy equation (1)
taking y and z coordinates,
$\dfrac{-3-1}{\lambda-1}=\dfrac{-4+2}{-1+2}$
$\lambda=3$

Now equation (1) becomes
$\dfrac{x-0}{3-0}=\dfrac{y-1}{3-1}=\dfrac{z+2}{-1+2}.........(2)$

Option (A) (12,9,2) satisfy equation (2).

Therefore option (A) is correct.