Tag: maths

Questions Related to maths

If $A$ is $(2, 4, 5),$ and $B$ is $(-7, -2, 8)$, then which of the following is collinear with$A$ and $B$ is

direction cosines and direction ratios three dimensional geometry maths

  1. $(1, 2, 6)$

  2. $(2, -1,6)$

  3. $(-1, 2, 6)$

  4. $(2, 6, -1)$

Show answer
Correct Option: C
Explanation:

Line passing through $A(2,4,5)$ and $B(-7,-2,8)$ is
$L: \dfrac{x-2}{9} = \dfrac{y-4}{6} = \dfrac{z-5}{-3}$
Lets suppose $O$ is collinear with $A$ and $B$, then $L = k$
$\therefore$ Coordinates of $O$ are $(2+9k,4+6k,5-3k)$
Looking at options, only option C satisfies the coordinates of $O$ for $k = -\dfrac{1}{3}$

A point $P$ lies on a line whose ends are $A(1,2,3)$ and $B(2,10,1).$ If $z$ component of $P$ is $7,$ then the coordinates of $P$ are

direction cosines and direction ratios three dimensional geometry maths

  1. $(-1,-14,7)$

  2. $(1,-14,7)$

  3. $(-1,14,7)$

  4. $(1,14,7)$

Show answer
Correct Option: A
Explanation:

The equation of line passing through $A$ and $B$ is $\displaystyle\frac { x-1 }{ 2-1 } =\frac { y-2 }{ 10-2 } =\frac { z-3 }{ 1-3 } \Rightarrow \frac { x-1 }{ 1 } =\frac { y-2 }{ 8 } =\frac { z-3 }{ -2 } =r$

Substitute $z=7$, we get
$\Rightarrow \displaystyle \frac { x-1 }{ 1 } =\frac { y-2 }{ 8 } =\frac { 7-3 }{ -2 } =-2$
Solve it to get $x=1-2=-1$, $y=2-16=-14$
So $P:$ $\left( -1,-14,7 \right) $

The vectors $\bar {a}=x\hat {i}-2\hat {j}+5\hat {k}$ and $\bar {b}=\hat {i}+y\hat {j}-z\hat {k}$are collinear if 

direction cosines and direction ratios three dimensional geometry maths

  1. $x=1$, $y=-2$, $z=-5$

  2. $x=1/2$, $y=-4$, $z=-10$

  3. $x=-1/2$, $y=4$, $z=-10$

  4. $x=-1$, $y=2$, $z=5$

Show answer
Correct Option: A
Explanation:
we have, $\vec{a} =x\hat{i}-x\hat{j}+5\hat{k}$ and $\vec{b} =\hat{i}+y\hat{j}-z\hat{k}$ 
Now, to have collinearity,
both vectors will be identical.
$\therefore x=1, y=-2,$ and $z=-5$ Ans

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?

direction cosines and direction ratios three dimensional geometry maths

  1. $2$

  2. $4$

  3. $6$

  4. $8$

Show answer
Correct Option: B
Explanation:

$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\, $

Three points whose position vectors are $\overrightarrow{a}$, $\overrightarrow{b}$, $\overrightarrow{c}$ will be collinear if

direction cosines and direction ratios three dimensional geometry maths

  1. $\lambda \overrightarrow{a}+\mu \overrightarrow{b}=\left ( \lambda +\mu \right )\overrightarrow{c}$

  2. $\overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{b}\times \overrightarrow{c}+\overrightarrow{c}\times \overrightarrow{a}=\overrightarrow{0}$

  3. $\begin{bmatrix}

    \overrightarrow{a} & \overrightarrow{b} & \overrightarrow{c}

    \end{bmatrix}=0$

  4. None of these

Show answer
Correct Option: A,B
Explanation:

If $\vec{a}$, $\vec{b}$ and $\vec{c}$ are collinear vectors, therefore they lie on the same line. Hence they are parallel. Hence there cross products will be 0.
Therefore
$(\vec{a}\times\vec{b})=(\vec{c}\times\vec{b})=(\vec{a}\times\vec{c})=0$
And application of the section formula gives us
$\vec{c}=\dfrac{\lambda \vec{a}+\mu\vec{b}}{\lambda+\mu}$
Or
$\vec{c}(\lambda+\mu)=\lambda \vec{a}+\mu\vec{b}$

Assertion ($A$): 

Three points with position vectors $\vec{a},\vec{b},\ \vec{c}$ are collinear if $\vec{a}\times\vec{b}+\vec{b}\times\vec{c}+\vec{c}\times\vec{a}=\vec{0}$
Reason ($R$):
Three points ${A}, {B},\ {C}$ are collinear if $\vec{AB}={t}\ \vec{BC}$, where ${t}$ is a scalar quantity.

direction cosines and direction ratios three dimensional geometry maths

  1. Both $A$ and $R$ are individually true and $R$ is the correct explanation of $A$.

  2. Both $A$ and $R$ are individually true and $R$ is NOT the correct explanation of $A$.

  3. $A$ is true but $R$ is false.

  4. $A$ is false but $R$ is true.

Show answer
Correct Option: A
Explanation:
Let the position vectors of $A, B, C$ be $\vec a, \vec b, \vec c$ respectively.
Given, $\vec{AB} = t\ \vec{BC}$
$\Rightarrow \vec{AB}\times\vec{BC} = 0$
$\Rightarrow (\vec b - \vec a)\times(\vec c - \vec b) = 0$
$\Rightarrow (\vec a - \vec b)\times(\vec b - \vec c) = 0$
$\Rightarrow \vec a\times \vec b + \vec b\times \vec c + \vec c \times \vec a = 0$
Hence, $\vec a, \vec b, \vec c$ are collinear.

Hence, option A.

Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three non-zero vectors, no two of which are collinear. If the vector $\vec{a}+2\vec{b}$ is collinear with $\vec{c}$ and $\vec{b}+3\vec{c}$ is collinear with $\vec{a}$, then $\vec{a}+2\vec{b}+6\vec{c}$ is equal to.

direction cosines and direction ratios three dimensional geometry maths

  1. $\lambda \vec{a}$

  2. $\lambda \vec{b}$

  3. $\lambda \vec{c}$

  4. $\vec{0}$

Show answer
Correct Option: D
Explanation:

Since, $\vec{a}+2\vec{b}$ is collinear with $\vec{c}$.
$\therefore \vec{a}+2\vec{b}=x\vec{c}, x\epsilon R$ and $\vec{b}+3\vec{c}$ is collinear with $\vec{a}$.
$\therefore \vec{b}+3\vec{c}=y\vec{a}, y\epsilon R$
$\Rightarrow \vec{a}+2\vec{b}+6\vec{c}=(1+2y)\vec{a}$
Also, $\vec{a}+2\vec{b}+6\vec{c}=(x+6)\vec{c}$
$\therefore (x+6)\vec{c}=(1+2y)\vec{a}$
$\Rightarrow x+6=0$
and $1+2y=0$
$\Rightarrow x=-6$ and $y=-1/2$
$\therefore \vec{a}+2\vec{b}+6\vec{c}=\vec{0}$

If $A$ , $B$ and $C$ are three collinear points, where $A= i + 8 j - 5k $, $ B  = 6i-2j$ and $C= 9i + 4j - 3 k$, then $B$ divides $AC$ in the ratio of :

direction cosines and direction ratios three dimensional geometry maths

  1. $\dfrac{5}{7}$

  2. $\dfrac{5}{3}$

  3. $\dfrac{2}{3}$

  4. None of these

Show answer
Correct Option: B
Explanation:

Given points are $A(i+8j-5k)$ , $B(6i-2j)$ and $C(9i+4j-3k)$

$AB = -5i+10j-5k$ and $CB = 3i+6j-3k$
$\Rightarrow \dfrac{|AB|}{|CB|}=\dfrac{\sqrt{150}}{\sqrt{54}}=\dfrac{5\sqrt6}{3\sqrt6}=\dfrac{5}{3}$
Therefore correct option is $B$

If the points $a(cos \alpha + i sin \alpha)$ , $b(cos \beta + i sin \beta)$ and $c(cos \gamma + isin \gamma)$ are collinear then the value of $|z|$ is:  
( where ${z = bc  \ sin(\beta-\gamma) + ca \ sin(\gamma-\alpha) + ab \ sin(\alpha - \beta) + 3i -4k}$ )

direction cosines and direction ratios three dimensional geometry maths

  1. $2$

  2. $5$

  3. $1$

  4. None of these.

Show answer
Correct Option: B
Explanation:
Given $a=cos\alpha+i\sin\alpha=e^{i\alpha}$ , $b=cos\beta+isin\beta=e^{i\beta}$ and $c=cos\gamma+isin\gamma=e^{i\gamma}$
Consider $bcsin(\beta-\gamma)=e^{i(\beta+\gamma)}sin(\beta-\gamma)=\frac{1}{2i}e^{i(\beta+\gamma)}(e^{i(\beta-\gamma)}-e^{-i(\beta-\gamma)}) = \frac{1}{2i}(e^{i(2\beta)}-e^{i(2\gamma)})$
Similarly we get $casin(\gamma-\alpha) = \frac{1}{2i}(e^{i(2\gamma)}-e^{i(2\alpha)})$ and $absin(\alpha-\beta) = \frac{1}{2i}(e^{i(2\alpha)}-e^{i(2\beta)})$
Therefore we get $bcsin(\beta-\gamma)+casin(\gamma-\alpha)+absin(\alpha-\beta)=0$
So we get $z=3i-4i$
$\Rightarrow |z|=5$

Three points $A(\bar a),B(\bar b),C(\bar c)$ are collinear if and only if?

direction cosines and direction ratios three dimensional geometry maths

  1. $(\bar b - \bar a) \times (\bar c-\bar a)=0$

  2. $(\bar b - \bar a) \times (\bar c-\bar a)=1$

  3. $(\bar b - \bar a) \cdot (\bar c-\bar a)=0$

  4. $(\bar b - \bar a) \cdot (\bar c-\bar a)=1$

Show answer
Correct Option: A
Explanation:

Consider the given points $A(\bar{a}),B(\bar{b}),C(\bar{c})$.
These three points determine two vectors $\vec{AB}$ and $\vec{AC}$

We know that, "two vectors $a,b$ are collinear if and only if $\vec{a} \times \vec{b}=0$".

Therefore two vectors $\vec{AB}$ and $\vec{AC}$ are collinear if and only if $\vec{AB} \times \vec{AC}=0$

$\vec{AB}=\vec{OB}-\vec{OA}=\bar{b}-\bar{a}$ and $\vec{AC}=\vec{OC}-\vec{OA}=\bar{c}-\bar{a}$

We have, two vectors $\vec{AB}$ and $\vec{AC}$ are collinear if and only if $\vec{AB} \times \vec{AC}=0$

$ \Rightarrow$ two vectors $\vec{AB}$ and $\vec{AC}$ are collinear if and only if $(\bar{b}-\bar{a}) \times (\bar{c}-\bar{a})=0$

Since the three points determine two vectors $\vec{AB}$ and $\vec{AC}$, we conclude that

Three points $A(\bar{a}),B(\bar{b}),C(\bar{c})$ are collinear if and only if $(\bar{b}-\bar{a}) \times (\bar{c}-\bar{a})=0$.