MATHS 3 Marks Question

$\overrightarrow r = 3\overrightarrow a + 5\overrightarrow b \\ $
Also , mid point of the line segment RQ is :  $= \frac{{3\overrightarrow a + 5\overrightarrow b + \overrightarrow a - 3\overrightarrow b }}{2} = \frac{{4\overrightarrow a + 2\overrightarrow b }}{2} = 2\overrightarrow a + \overrightarrow b$ ,which is the position vector of point P.Therefore , P is the mid point of line segment RQ.

Let vector $\vec c$ be the resultant vector of $\vec a$ and $\vec b$

$\therefore \vec c = \vec a + \vec b $ $= 2\hat i + 3\hat j - \hat k + \hat i - 2\hat j + \hat k$

$ = 3\hat i + \hat j + 0\hat k$

$\therefore$ Required vector of magnitude 5 units and parallel (or collinear) to resultant vector $\vec c = \vec a + \vec b$ is

$5\hat c = 5\frac{{\vec c}}{{\left| {\vec c} \right|}} = 5\left( {\frac{{3\hat i + \hat j + 0\hat k}}{{\sqrt {9 + 1 + 0} }}} \right)$

$ = \frac{5}{{\sqrt {10} }}\left( {3\hat i + \hat j} \right)$

$= \frac{5}{{\sqrt {10} }} \times \frac{{\sqrt {10} }}{{\sqrt {10} }}\left( {3\hat i + \hat j} \right)$

$ = \frac{{5\sqrt {10} }}{{10}}\left( {3\hat i + \hat j} \right)$

$ \Rightarrow 5\hat c = \frac{{\sqrt {10} }}{2}\left( {3\vec i + \vec j} \right)$

$= \frac{{3\sqrt {10} }}{2}\hat i + \frac{{\sqrt {10} }}{2}\hat j$

$\vec{AB}=\hat i +2\hat j+3\hat k$, $\vec{AC}=4\hat j+3\hat k,$

$\therefore \overrightarrow{AB}\times \overrightarrow{AC}=\left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k} \\ 1&2&3 \\ 0&4&3 \end{array}} \right|=-6\hat i-3\hat j+4\hat k$

$\Rightarrow |\overrightarrow{AB}\times\vec{AC}|=\sqrt{61}$

$\Rightarrow\frac{1}{2} |\overrightarrow{AB}\times\vec{AC}|=\frac{1}{2}\sqrt{61}$
Therefore, the area of triangle ABC is =${1 \over 2}\sqrt {61} $

$\Rightarrow \vec a.\vec a - \vec a.\vec b + \vec b.\vec a - \vec b.\vec b = 8$

$ \Rightarrow {\left| {\vec a} \right|^2} - \vec a.\vec b + \vec a.\vec b - {\left| {\vec b} \right|^2} = 8$

$ \Rightarrow {\left| {\vec a} \right|^2} - \left| {\vec b} \right|^2 = 8$ ...(ii)

Putting ${\left| {\vec a} \right|^2} = 64\left| {\vec b} \right||^2$ in eq. (ii),

$64{\left| {\vec b} \right|^2} - \left| {\vec b} \right|^2 = 8$

$ \Rightarrow \left( {64 - 1} \right){\left| {\vec b} \right|^2} = 8$

$\Rightarrow \left( {63} \right){\left| {\vec b} \right|^2} = 8$

$ \Rightarrow {\left| {\vec b} \right|^2} = \frac{8}{{63}}$

$ \Rightarrow {\left| {\vec b} \right|^2} = \sqrt {\frac{8}{{63}}} = \frac{{2\sqrt 2 }}{{3\sqrt 7 }}$

Putting ${\left| {\vec b} \right|^2} = \frac{{2\sqrt 2 }}{{3\sqrt 7 }}$ in eq (i),

${\left| {\vec a} \right|^2} = 8\left( {\frac{{2\sqrt 2 }}{{3\sqrt 7 }}} \right) = \frac{{16}}{3}\sqrt {\frac{2}{7}} $

$ \Rightarrow \frac{{\overrightarrow a .\overrightarrow b }}{{\left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|}} = \cos \theta \Rightarrow \frac{{10}}{{14}} = \cos \theta $

$\Rightarrow \cos \theta = \frac{5}{7} \Rightarrow \theta = {\cos ^{ - 1}}\frac{5}{7} \\\\\\ $

$=\frac{(2\hat i+2\hat j+3\hat k).(\hat i+\hat j+2\hat k)}{\sqrt{17}\sqrt6}$

$\Rightarrow cos\theta=\frac{10}{\sqrt{102}}$

$\Rightarrow \angle ABC=cos^{-1}(\frac{10}{\sqrt{102}})$

Now $\vec a + \lambda \vec b = 2\hat i + 2\hat j + 3\hat k + \lambda \left( { - \hat i + 2\hat j + \hat k} \right)$=

$ = 2\hat i + 2\hat j + 3\hat k - \lambda \hat i + 2\lambda \hat j + \lambda \hat k$

$\Rightarrow \vec a + \lambda \vec b = \left( {2 - \lambda } \right)\hat i + \left( {2 + 2\lambda } \right)\hat j + \left( {3 + \lambda } \right)\hat k$

Again, $\vec c = 3\hat i + \hat j = 3\hat i + \hat j + 0\hat k$

Since, $\left( {\vec a + \lambda \vec b} \right)$ is perpendicular to $\vec c$ therefore, $\left( {\vec a + \lambda \vec b} \right).\vec c = 0$

$\Rightarrow 6 - 3\lambda + 2 + 2\lambda = 0 \Rightarrow - \lambda + 8 = 0$

$ \Rightarrow - \lambda = - 8 \Rightarrow \lambda = 8$

Let us find angle ${\theta _1}$ (say) between vector $\vec a$ and OX $$

$\Rightarrow \cos {\theta _1} = \frac{{\vec a.\hat i}}{{\left| {\vec a} \right|.\left| {\hat i} \right|}} $ $= \frac{{\left( {i + j + k} \right).\left( {\hat i + 0\hat j + 0\hat k} \right)}}{{\left| {\hat i + \hat j + \hat k} \right|\left| {\hat i + 0\hat j + 0\hat k} \right|}} $ $= \frac{{1\left( 1 \right) + 1\left( 0 \right) + 1\left( 0 \right)}}{{\sqrt {1 + 1 + 1} .\sqrt {1 + 0 + 0} }} = \frac{1}{{\sqrt 3 }}$

$ \Rightarrow {\theta _1} = {\cos ^{ - 1}}\frac{1}{{\sqrt 3 }}$

Similarly angle ${\theta _2}$ (say) between vector $\vec a$ and OY $$ is ${\cos ^{ - 1}}\frac{1}{{\sqrt 3 }}$

And angle ${\theta _3}$ (say) between vector $\vec a$ and OZ $$ is ${\cos ^{ - 1}}\frac{1}{{\sqrt 3 }}$

$\therefore {\theta _1} = {\theta _2} = {\theta _3}$

$\therefore$ Vector $\vec a = \hat i + \hat j + \hat k$ is equally inclined to OX, OY and OZ.

$\therefore \left| {\vec a} \right| = \sqrt {{1^2} + {1^2} + {1^2}} = \sqrt {1 + 1 + 1} = \sqrt 3 $

And $\vec b = 2\hat i - 7\hat j - 3\hat k$

$\therefore \left| {\vec b} \right| = \sqrt {{{\left( 2 \right)}^2} + {{\left( { - 7} \right)}^2} + {{\left( { - 3} \right)}^2}} $ $= \sqrt {4 + 49 + 9} = \sqrt {62} $

Also $\vec c = \frac{1}{{\sqrt 3 }}\hat i + \frac{1}{{\sqrt 3 }}\hat j - \frac{1}{{\sqrt 3 }}\hat k$

$\therefore \left| {\vec c} \right| = \sqrt {{{\left( {\frac{1}{{\sqrt 3 }}} \right)}^2} + {{\left( {\frac{1}{{\sqrt 3 }}} \right)}^2} + {{\left( {\frac{{ - 1}}{{\sqrt 3 }}} \right)}^2}} $ $= \sqrt {\frac{1}{3} + \frac{1}{3} + \frac{1}{3}} = 1$

The area of a parallelogram with $\vec a$ and $\vec b$ as its adjacent sides is given by $|\vec{a} \times \vec{b}|$.
Now, $\vec{a} \times \vec{b}=\left|\begin{array}{ccc} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ {3} & {1} & {4} \\ {1} & {-1} & {1} \end{array}\right|=5 \hat{i}+\hat{j}-4 \hat{k}$ 
Therefore $|\vec{a} \times \vec{b}|=\sqrt{25+1+16}=\sqrt{42}$ 
Hence, the required area is $\sqrt{42}$.

We have $\vec{\mathrm{AB}}=\hat{j}+2 \hat{k}$ and $\vec{\mathrm{AC}}=\hat{i}+2 \hat{j}$.
The area of the given triangle is $\frac{1}{2}|\vec{\mathrm{AB}} \times \vec{\mathrm{AC}}|$
Now, $\vec{\mathrm{AB}} \times \vec{\mathrm{AC}}=\left|\begin{array}{lll} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ {0} & {1} & {2} \\ {1} & {2} & {0} \end{array}\right|=-4 \hat{i}+2 \hat{j}-\hat{k}$ 
Therefore, $|\vec{\mathrm{AB}} \times \vec{\mathrm{AC}}|=\sqrt{16+4+1}=\sqrt{21}$ 
Thus, the required area is $\frac{1}{2} \sqrt{21}$

$B ( \hat { i } + 2 \hat { j } + 3 \hat { k } ),$ and 

$C ( 7 \hat { i} - \hat { k } ).$
Consider, $\vec { A B } = ( \hat {i } + 2 \hat { j } + 3 \hat { k } ) - ( - 2 \hat { i } + 3 \hat {j} + 5 \hat { k } )$$= 3 \hat { i } - \hat { j } - 2 \hat { k }$
Similarly, $\vec { B C } = ( { 7} \hat { i } - \hat { k } ) - ( \hat { i} + 2 \hat { j } + 3 \hat { k } )$$= 6 \hat {i} - 2 \hat {j} - 4 \hat { k }$
Similarly, $\vec { A C } = ( 7 \hat { i } - \hat { k } ) - ( - 2 \hat {i} + 3 \hat { j } + 5 \hat { k } )$$= 9 \hat {i } - 3 \hat { j } - 6 \hat { k }$
Now, consider, $\vec { A B } + \vec { B C } = ( 3 \hat { i } - \hat { j } - 2 \hat { k } )$$+ ( 6 \hat { i } - 2 \hat {j } - 4 \hat { k } ) = 9 \hat { i} - 3 \hat { j } - 6 \hat { k } = \vec { A C }$
$\therefore$ The points A, B and C are collinear.