Mathematics M.C.Q (1 Marks)

Here, mid points of $\overrightarrow {AC} $ and $\overrightarrow {BD} $ coincide, where   

$\overrightarrow {AC} $ and $\overrightarrow {BD} $ are diagonals. In addition, we know that diagonals of a parallelogram bisect each other.
Hence quadrilateral is parallelogram.

$⇒$ $m = \frac{1}{{|a|}}$.

or $\overrightarrow {AB} = \overrightarrow {CQ} $.

Hence it is a parallelogram.

$w = b + a$…..$(ii)$

We have, $x = v + w = a + 2b + c$.

$ = (i + 2j + 3k) - (2i + 4j - 5k) = - i - 2j + 8k$

Hence unit vector in the direction of $\overrightarrow {BD} $ is

$\frac{{ - i - 2j + 8k}}{{| - i - 2j + 8k|}} = \frac{{ - i - 2j + 8k}}{{\sqrt {69} }}.$

$\therefore \,2x + y - 3z = - 2,$ $ - 3x - 2y + z = 3$ and $y + 2z = - 1$

Solving these, we get $x = 0,$ $y = - \frac{7}{5},$ $z = \frac{1}{5}$

 $ - 2a + 3b - c = \frac{{( - 7q + r)}}{5}.$

Trick : Check alternates one by one

i.e., $(a)$ $p - 4q = - 2a + 5b - 4c$

$(b)$ $\frac{{ - 7q + r}}{5} = - 2a + 3b - c$.

$ = \lambda \,\overrightarrow {AD} + (\overrightarrow {AC} + \overrightarrow {CD} ) = \lambda \,\overrightarrow {AD} + \overrightarrow {AD} = (\lambda + 1)\overrightarrow {AD} .$

Therefore $p = \mu \overrightarrow {AD} \Rightarrow \mu = \lambda + 1.$

$\overrightarrow {DA} + \overrightarrow {DB} + \overrightarrow {DC} + \overrightarrow {AE} + \overrightarrow {BE} + \overrightarrow {CE} $

=$(\overrightarrow {DA} + \overrightarrow {AE} ) + (\overrightarrow {DB} + \overrightarrow {BE} ) + (\overrightarrow {DC} + \overrightarrow {CE} )$

= $\overrightarrow {DE} + \overrightarrow {DE} + \overrightarrow {DE} $=$3\overrightarrow {DE} .$

Resultant $ = (\overrightarrow {AC} + \overrightarrow {AD} + \overrightarrow {AE} ) + (\overrightarrow {CB} + \overrightarrow {DB} + \overrightarrow {EB} )$

$ = (\overrightarrow {AC} + \overrightarrow {CB} ) + (\overrightarrow {AD} + \overrightarrow {DB} ) + (\overrightarrow {AE} + \overrightarrow {EB} )$

$ = \overrightarrow {AB} + \overrightarrow {AB} + \overrightarrow {AB} = 3\overrightarrow {AB} $.

$\overrightarrow {O'B} = \overrightarrow {O'O} + \overrightarrow {OB} $

$\overrightarrow {O'C} = \overrightarrow {O'O} + \overrightarrow {OC} $

$ \Rightarrow \overrightarrow {O'A} + \overrightarrow {O'B} + \overrightarrow {O'C} $

$ = 3\overrightarrow {O'O} + \overrightarrow {OA} + \overrightarrow {OB} + \overrightarrow {OC} $

Since $\overrightarrow {OA} + \overrightarrow {OB} + \overrightarrow {OC} = \overrightarrow {OO'} = - \overrightarrow {O'O} $

$\therefore $ $\overrightarrow {O'A} + \overrightarrow {O'B} + \overrightarrow {O'C} = 2\overrightarrow {O'O} $.

Obviously, if $\overrightarrow {BC} $ is added to this system, then it will be $\overrightarrow {AC} + \overrightarrow {AB} + \overrightarrow {BC} = \overrightarrow {AC} + \overrightarrow {AC} = 2\overrightarrow {AC} .$

$\overrightarrow {BE}  + \overrightarrow {AF}  = \frac{{\overrightarrow {OA}  + \overrightarrow {OC} }}{2} - \overrightarrow {OB}  + \frac{{\overrightarrow {OB}  + \overrightarrow {OC} }}{2} - \overrightarrow {OA} $

$\overrightarrow {BE}  + \overrightarrow {AF}  = \overrightarrow {OC}  - \frac{{\overrightarrow {OA}  + \overrightarrow {OB} }}{2} = \overrightarrow {OC}  - \overrightarrow {OD}  = \overrightarrow {DC} $.

Now, $c.a = 0 \Rightarrow 2\lambda + \mu = 0 \Rightarrow \mu = - 2\lambda $

Therefore, $c = - \lambda i - \lambda j - 2\lambda k = (\sqrt 6 )( - \lambda )\left[ {\frac{{i + j + 2k}}{{\sqrt 6 }}} \right]$

Hence, unit vector $ = \frac{{(i + j + 2k)}}{{\sqrt 6 }}$.

The points are collinear, then $\overrightarrow {AB} = k\,(\overrightarrow {BC} )$

$ - 2j = k\{ (a - 1)i + (b + 1)\,j + ck\} $

On comparing, $k\,(a - 1) = 0$, $k(b + 1) = - 2,$ $kc = 0$.

Hence $c = 0,$ $a = 1$ and $b$ is arbitrary scalar.

$\overrightarrow {AB}  =  - 2b,$ $\overrightarrow {BC}  = (k + 1)b$

 $\forall \,\,k \in R \Rightarrow \overrightarrow {AB}  = \lambda \overrightarrow {BC} .$

Therefore, it is of the form $\overrightarrow {AB} = m\overrightarrow {AC} .$

Hence $A, B, C$  are collinear.

or $2i - 8j = k[(a - 12)i + 16j] \Rightarrow k = \frac{{ - 1}}{2}$

Also, $2 = k(a - 12) \Rightarrow a = 8.$

 So ,$\frac{{ - 1}}{{ - 2}} = \frac{{ - 4}}{{\lambda - 2}}$, $\lambda - 2 = - 8$ or $\lambda = - 6.$

$ \Rightarrow (x - 2)a + b = \lambda (2x + 1)a - \lambda b$

or $[(x - 2) - \lambda (2x + 1)]a + (\lambda + 1)b = 0$

$(x - 2) - \lambda (2x + 1) = 0,\lambda + 1 = 0$

$(\because \,a,b$ are linearly independent)

$ \Rightarrow x - 2 + 2x + 1 = 0 \Rightarrow x = \frac{1}{3}.$

Hence $\overrightarrow {PQ} \,||\,\overrightarrow {RS} $.

$ \Rightarrow ( - 2i + mj) = \lambda (i - j)$

Comparison of coefficient, we get

$ \Rightarrow \lambda = - 2$ and $ - \lambda = m.$ So, $m = 2$.

$5i - 2k = \frac{{\lambda (11i + 3j + 7k) + i - 2j - 8k}}{{\lambda + 1}}$

$ \Rightarrow 3\lambda - 2 = 0$ $ \Rightarrow \lambda = \frac{2}{3}$ $ i.e., ratio = 2 : 3.$

Because otherwise one will be a scalar multiple of the other and hence collinear which is a contradiction.

==> $2i + (4 - x)j + 4k = \lambda \,[(y - 3)i - 6j - 12k]$

$ \Rightarrow \,\,\,2 = (y - 3)\lambda $ .....$(i)$

and $4 - x = - 6\,\lambda $ .....$(ii)$

==> $4 = - 12\,\lambda \,\,\,\,\, \Rightarrow \,\,\lambda = \frac{{ - 1}}{3}$

By $(i),$ $y = - 3$ and by $(ii),$ $x = 2$; $\therefore \,\,\,(x,\,y) = (2,\, - 3).$

==> $i - j = m{\rm{ }}( - 2i + kj)$

==> $i - j = - 2mi + kmj$

==> $ - 2m = 1,\,km = - 1$

$m = - \frac{1}{2},$ So $k = 2$.

Unit vector $ = \frac{{(2 + b)i + 6j - 2k}}{{\sqrt {{b^2} + 4b + 44} }}$

According to the condition, $1 = \frac{{(2 + b) + 6 - 2}}{{\sqrt {{b^2} + 4b + 44} }}$

$ \Rightarrow {b^2} + 4b + 44 = {b^2} + 12b + 36$ $ \Rightarrow 8b = 8 \Rightarrow b = 1.$

$2q = (2y - 4x + 4)\,a + (4x - 6y - 2)\,b$

On comparing, we get $3x + 12y = 2y - 4x + 4$

==> $7x + 10y = 4$ …..$(i)$

and $2x + 9y = - 5$ …..$(ii)$

On solving equations, we get $x = 2,\,\, - 1.$

$\overrightarrow {{\rm{AD}}}  = \frac{{8{\rm{i}} + 6{\rm{k}}}}{2} = 4{\rm{i}} + 3{\rm{k}}$

$\therefore $ Length of median $=|\overrightarrow{\mathrm{AD}}|=\sqrt{16+9}=5$ $unit.$

Angle between $a$ and $a + b + c$ is $\cos \theta = \frac{{a.(a + b + c)}}{{|a||a + b + c|}}$ .....$(i)$

Now $|a| = |b| = |c| = a$

$|a + b + c{|^2} = {a^2} + {b^2} + {c^2} + 2\,a\,.\,b + 2\,b\,.\,c + 2\,c\,.\,a$

$ = {a^2} + {a^2} + {a^2} + 0 + 0 + 0$

$\Rightarrow  |a + b + c{|^2} = 3{a^2} \Rightarrow |a + b + c| = \sqrt 3 a$

Putting this value in $(i),$ we get $\theta = {\cos ^{ - 1}}\frac{1}{{\sqrt 3 }}.$

Therefore $|a|\, = \,|b|\, = \,|c|\, = 1$ and $a.b = b.c = c.a = 0$.

We know that

$|a + b + c{|^2} = (a + b + c)\,.\,(a + b + c) = \,\,|a{|^2} + |b{|^2}$

$ + |c{|^2} + 2(a\,.\,b\,\, + b\,.\,c\, + c\,.\,a) = 1 + 1 + 1 + 0 = 3$

or $|a + b + c|\, = \sqrt 3 .$

${\cos ^{ - 1}}\left\{ {\frac{{(i + j + k)\,.\,i}}{{|i + j + k|\,\,|i|}}} \right\} \Rightarrow \alpha = {\cos ^{ - 1}}\left( {\frac{1}{{\sqrt 3 }}} \right)$

Similarly angle between $i + j + k$ and $j$ is $\beta = {\cos ^{ - 1}}\left( {\frac{1}{{\sqrt 3 }}} \right)$ and between $i + j + k$ and

$k$ is $\gamma = {\cos ^{ - 1}}\left( {\frac{1}{{\sqrt 3 }}} \right)\,.$

Hence $\alpha = \beta = \gamma .$

Since given that $\cos 30^\circ = \frac{{(yj + zk)\,.\,(yj)}}{{|yj + zk|\,\,|yj|}}$

$ \Rightarrow \frac{{{y^2}}}{{\left( {\sqrt {{y^2} + {z^2}} } \right)\,y}} = \frac{{\sqrt 3 }}{2} \Rightarrow y = \frac{{\sqrt 3 }}{2}$,

$(\,\sqrt {{y^2} + {z^2}}  = 1$by $(i)$)

Similarly, $\cos 60^\circ = \frac{{(yj + zk)\,.\,zk}}{{|yj + zk|\,\,|zk|}} \Rightarrow z = \frac{1}{2}$

Hence the components of unit vector are $0,\,\,\frac{{\sqrt 3 }}{2},\,\,\frac{1}{2}.$

Trick : Since the vector lies in $yz - $plane, so it will be either $0i + \frac{{\sqrt 3 }}{2}j + \frac{1}{2}k$ or $0i + \frac{1}{2}j + \frac{{\sqrt 3 }}{2}k.$

But the vector $\frac{{\sqrt 3 }}{2}j + \frac{1}{2}k$ makes angle $30^\circ $with $y - $ axis and that of $60^\circ $ with $z-$ axis.

$ \Rightarrow 0 = 7 + 14\lambda \,\, \Rightarrow \lambda \,\, = - \frac{1}{2}$

Therefore, $r = - \frac{1}{2}(i + 5j - 4k).$

$(AB)\,(AC)\cos \theta + (BC)\,(BA)\cos ({90^o} - \theta ) + 0$

=$AB(AC\cos \theta + BC\sin \theta ) = AB\left( {\frac{{{{(AC)}^2}}}{{AB}} + \frac{{{{(BC)}^2}}}{{AB}}} \right)$

$ = A{C^2} + B{C^2} = A{B^2} = {p^2}$.

$\overrightarrow {AC} = \overrightarrow {AB} + \overrightarrow {BC} = a + b$ or $\overrightarrow {CA} = - (a + b)$

$\overrightarrow {BD} = \overrightarrow {BC} + \overrightarrow {CD} = b + c$

Therefore, $\overrightarrow {AB} \,.\,\overrightarrow {CD} + \overrightarrow {BC} \,.\,\overrightarrow {AD} + \overrightarrow {CA} \,.\,\overrightarrow {BD} $

$ = a\,.\,c + b\,.(a + b + c) + ( - a - b)\,.\,(b + c)$

$ = a\,.\,c + b\,.\,a + b\,.\,b + b\,.\,c - a\,.\,b - a\,.\,c - b\,.\,b - b\,.\,c$

$ = 0$.

and $|a - c|\, = \,|b - c|$ $ \Rightarrow $ $\,|a - c{|^2}\, = \,|b - c{|^2}$

$\therefore  a + b = 2c$

Therefore, $(b - a).\,\left( {c - \frac{{a + b}}{2}} \right) = 0.$

Then $a\,.\,i = (xi + yj + zk)\,.\,i = x$ and $a\,.\,(i + j) = x + y$ and $a\,.\,(i + j + k) = x + y + z$

Given that $x = x + y = x + y + z$

Now $x = x + y\,\,\, \Rightarrow y = 0$  and $x + y = x + y + z\,\, \Rightarrow \,\,z = 0$

Hence $x = 1$; $\therefore \,\,a = i$.