Vector Algebra - Class 12 Engineering Maths - Extra Questions

The position vector of mid-point $$R$$ of the vector joining points $$P (2, 3, 4)$$ and $$Q (4, 1, -2)$$ is given by,
$$\vec {OR}=\dfrac {(2\hat {i}+3\hat {j}+4\hat {k})+(4\hat {i}+\hat {j}-2\hat {k})}{2}=\dfrac {(2+4)\hat {i}+(3+1)\hat {j}+(4-2)\hat {k}}{2}$$
$$=\dfrac {6\hat {i}+4\hat {j}+2\hat {k}}{2}=3\hat {i}+2\hat {j}+\hat {k}$$

Given that:

Let the ratio be $$\lambda :1$$
Position vector of point $$(C)$$
$$\vec { c } =\cfrac { \lambda \vec { a } +\vec { b }  }{ \lambda +1 } $$
Put the value of $$\overrightarrow { c } $$ in $$\quad 2\vec { a } +3\vec { b } -5\vec { c } =0$$
$$2\vec { a } +3\vec { b } -5\times \left( \cfrac { \lambda \vec { a } +\vec { b }  }{ \lambda +1 }  \right) =0$$
$$\Rightarrow \cfrac { \left( \lambda +1 \right) \left( 2\vec { a } +3\vec { b }  \right) -5\lambda \vec { a } -5\vec { b }  }{ \left( \lambda +1 \right)  } =0\quad $$
$$\Rightarrow 2\vec { a } \lambda +3\vec { b } \lambda +2\vec { a } +3\vec { b } -5\lambda \vec { a } -5\vec { b } =0$$
$$\Rightarrow 3\vec { b } \lambda -3\vec { a } \lambda +2\vec { a } -2\vec { b } =0\quad $$
$$\Rightarrow 3\lambda (\vec { b } -\vec { a } )=2\vec { b } -2\vec { a } $$
$$\Rightarrow 3\lambda =\cfrac { 2(\vec { b } -\vec { a } ) }{ \left( \vec { b } -\vec { a }  \right)  } \quad \Rightarrow \cfrac { \lambda  }{ 1 } =\cfrac { 3 }{ 2 } $$
hence the point $$C$$ divides line segment $$AB$$ in the ratio $$2:3$$.

$${\left(A+B\right)}^{2}=\left(A+B\right)\left(A+B\right)$$

Area of parallelogram is magnitude of cross product of two vectors

$$|\bar a|=3\\|\bar b|=4\\|\bar c|=5$$

a. Resultant is maximum when both vectors act in same direction. $$R_{max} = A + B = 7 + 3 = 10 \ N$$

The position vector of point $$R$$ dividing the line segment joining two points $$P$$ and $$Q$$ in the ratio $$m:n$$ is given by:
i. Internally:
$$\dfrac {m\vec {Q}+n\vec {P}}{m+n}$$
ii. Externally:
$$\dfrac {m\vec {Q}-n\vec {P}}{m-n}$$
Position vectors of $$P$$ and $$Q$$ are given as:
$$\vec {OP}=\hat {i}+2\hat {j}-\hat {k}$$ and $$\vec {OQ}=-\hat {i}+\hat {j}+\hat {k}$$
(i) The position vector of point $$R$$ which divides the line joining two points $$P$$ and $$Q$$ internally in the ratio $$2:1$$ is given by,
$$\vec {OR}=\dfrac {2(-\hat {i}+\hat {j}+\hat {k})+1(\hat {i}+2\hat {j}-\hat {k})}{2+1}=\dfrac {-2\hat {i}+2\hat {j}+\hat {k}+(\hat {i}+2\hat {j}-\hat {k})}{3}$$
$$=\dfrac {-\hat {i}+4\hat {j}+\hat {k}}{3}=-\dfrac {1}{3}\hat {i}+\dfrac {4}{3}\hat {j}+\dfrac {1}{3}\hat {k}$$
(ii) The position vector of point $$R$$ which divides the line joining two points $$P$$ and $$Q$$ externally in the ratio $$2:1$$ is given by,
$$\vec {OR}=\dfrac {2(-\hat {i}+\hat {j}+\hat {k})-1(\hat {i}+2\hat {j}-\hat {k})}{2-1}=(-2\hat {i}+2\hat {j}+2\hat {k})-(\hat {i}+2\hat {j}-\hat {k})$$
$$=-3\hat {i}+3\hat {k}$$

$$(\vec x+\vec y)\cdot (\vec x-\vec y)=63$$

$${ \left| \overline { R }  \right|  }^{ 2 }=\overline { R } .R\implies\quad { \left| \overline { R }  \right|  }^{ 2 }=(\overline { A } +\overline { B } ).(\overline { A } +\overline { B } )$$

$$\begin{array}{l} \vec { a } =5\hat { i } -2\hat { j } +5\hat { k }  \\  \\ \vec { b } =3\hat { i } +\hat { k }  \end{array}$$

$$\vec {OA}=\vec a$$

P.V of $$\vec{D}=\dfrac{2(\vec{(c)}+1(\vec{b})}{2+1}=\dfrac{2\vec{c}+\vec{b}}{3}$$

$$\left[2\hat{i}\hat{j}\hat{k}\right]+\left[\hat{i}\hat{k}\hat{j}\right]+\left[\hat{k}\hat{j}2\hat{i}\right]$$

$$ \therefore |\overrightarrow {a} \times \hat {i} |^2 + | \overrightarrow {a} \times  \hat {k}|^2 = a^2 sin^2 \alpha + a^2 sin \beta + a^2 sin^2 \gamma $$ 
$$ = a^2 [ sin^2 \alpha + sin^2 \beta  + sin^2 \gamma ] $$
$$ = a^2 [ 1- cos^2 \alpha +1 -cos^2 \beta +1 -cos^2 \gamma ] $$
$$= a^2 [ 3- (cos^2 \alpha  + cos^2 \beta +cos^2 g) ] $$
$$= a^2 (3- 1)          [ \because l^2 +m^2 +n^2 = 1 \Rightarrow cos^2 \alpha + cos^2 \beta + cos^2 \gamma = 1] $$
$$ = 2a^2 $$

$$ P.V. $$ of mid point

$$ \overline{AB}=-\overrightarrow{i}+3\overrightarrow{j}+5\bar{k},\quad \overrightarrow{BC}=-\overrightarrow{i}-2\overrightarrow{j}-6\bar{k},\overrightarrow{AC}=-2\overrightarrow{i}+\overrightarrow{j}-\bar{k} $$

$$(i) $$

Add the vectors $$\vec { a } =\hat { i } -2\hat { j } \\$$, $$\vec { b } =2\hat { i } -3\hat { j } \\$$ and $$\vec { c } =2\hat { i } +3\hat { k }$$  by writing them in the component form that is

Add the vectors 

$$\begin{matrix} \vec { a } =\frac { { 3\hat { i } +5\hat { j } -\hat { k }  } }{ { \sqrt { 35 }  } } ,\vec { b } =\frac { { 5\hat { i } -\hat { j } +10\hat { k }  } }{ { \sqrt { 126 }  } }  \\ Then,\, \, \, \left[ { \vec { a } +\vec { b } \cdot \vec { a } +2\vec { b } \vec { a } \times \vec { b }  } \right]  \\ \Rightarrow \vec { b } \cdot \vec { a } =\left( { \frac { { 3\hat { i } +5\hat { j } -\hat { k }  } }{ { \sqrt { 35 }  } }  } \right) \cdot \left( { \frac { { 5\hat { i } -\hat { j } +10\hat { k }  } }{ { \sqrt { 126 }  } }  } \right)  \\ \Rightarrow \vec { b } \cdot \vec { a } =0 \\ and\, \, 2\vec { b } \cdot \vec { a } \times \vec { b } =0 \\ \therefore \left[ { \vec { a } +\vec { b } \cdot \vec { a } +2\vec { b } \cdot \vec { a } \times \vec { b }  } \right] =\left[ { \frac { { 3\hat { i } +5\hat { j } -\hat { k }  } }{ { \sqrt { 35 }  } } +0+0 } \right]  \\ =\frac { { 3\hat { i }  } }{ { \sqrt { 35 }  } } +\frac { { 5\hat { j }  } }{ { \sqrt { 35 }  } } -\frac { { \hat { k }  } }{ { \sqrt { 35 }  } } \, \, \, \, \, \, Ans. \\  \end{matrix}$$

Taking A as origin let $$\overrightarrow{b} $$ and $$\overrightarrow {c}$$ be the postion vectors of $$B$$ and $$C$$, respectively,

The position vector of $$Q$$ is $$\dfrac{3 \overrightarrow {b}+\overrightarrow {c}}{4}$$ and that of $$p$$ is $$\dfrac{\overrightarrow {b}}{4}$$

if $$\dfrac{AR}{QR}=\dfrac{λ}{1}$$ then position vector of $$R = \lambda\left(\dfrac{3 \overrightarrow {b}+\overrightarrow {c}}{4}\right)$$    $$..(i)$$

if $$\dfrac{CR}{RP}=\dfrac{μ}{1}$$ then position vector of $$R = \dfrac{\mu \frac{\overrightarrow {b}}{4}+\overrightarrow {c}}{\mu +1}$$     $$...(ii)$$

comparing $$(i)$$ and $$(ii)$$, we have,

$$\dfrac {3 \lambda}{4}=\dfrac{\mu}{4(\mu+1)}$$ and $$\dfrac{\lambda}{4}=\dfrac{1}{\mu +1}$$

solving , $$\lambda=\dfrac{4}{13}$$ and $$\mu=12$$

Therefore, position , vector $$R$$ is $$\dfrac{3 \overrightarrow {b}+ \overrightarrow {c}}{13}$$

$$\dfrac{\triangle ABC} {\triangle BRC}=\dfrac{\left| \dfrac{3 \overrightarrow {b} + \overrightarrow {c}}{4}\right|}{\left| \dfrac{3 \overrightarrow {b}+\overrightarrow {c}}{4} - \left( \dfrac{3\overrightarrow {b}+\overrightarrow {c}}{13} \right) \right|}=\dfrac{13}{9}$$

Area of $$\triangle ABC$$ is $$\dfrac{13}{9}$$ units