Tag: direction cosines and direction ratios

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Questions Related to direction cosines and direction ratios

Multiple choice direction cosines and direction ratios three dimensional geometry maths

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 :

  1. $\dfrac{5}{7}$
  2. $\dfrac{5}{3}$
  3. $\dfrac{2}{3}$

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $B$

Multiple choice direction cosines and direction ratios three dimensional geometry maths

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 ${z = bc  \ sin(\beta-\gamma) + ca \ sin(\gamma-\alpha) + ab \ sin(\alpha - \beta) + 3i -4k}$ )

  1. $2$
  2. $5$
  3. $1$

  4. None of these.
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $z=3i-4i$
$\Rightarrow |z|=5$
Multiple choice direction cosines and direction ratios three dimensional geometry maths

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

  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. $(\bar b - \bar a) \cdot (\bar c-\bar a)=1$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $(\bar{b}-\bar{a}) \times (\bar{c}-\bar{a})=0$.