Mathematics M.C.Q (1 Marks)

So, $(a + \lambda b).(a - \lambda b) = 0$ ==> $|a{|^2} - {\lambda ^2}|b{|^2} = 0$

or ${\lambda ^2} = \frac{{|a{|^2}}}{{|b{|^2}}} \Rightarrow {\lambda ^2} = \frac{9}{{16}}$ or $\lambda = \pm \frac{3}{4}$, 

$[\because \,|a| = 3,|b| = 4]$

Using dot product, $(a + b).(a + b) = c.c$

$ \Rightarrow a.a + b.b + 2a.b = c.c$

$ \Rightarrow \,|a|.|a|\cos 0^\circ + |b|.|b|\cos 0^\circ + 2|a|.|b|\cos \alpha $

$ = \,|c|.|c|\cos 0^\circ $,  $(\because \,\,\,|a|\, = \,|b|\, = \,|c|\, = 1)$

$ \Rightarrow 1 + 1 + 2\cos \alpha = 1 \Rightarrow \cos \alpha = - \frac{1}{2} \Rightarrow \alpha = \frac{{2\pi }}{3}$.

Also $|a + b{|^2} = {1^2} \Rightarrow 1 + 1 + 2\cos \theta = 1$$ \Rightarrow \theta = 120^\circ $

$\therefore \,\,\,|a - b{|^2} = 1 + 1 - 2\cos \theta = 3$$ \Rightarrow \,|a - b|\, = \sqrt 3 .$

$ \Rightarrow \sqrt 2  \times \sqrt 2 \,\sin \frac{\theta }{2} = 2\sin \frac{\theta }{2}$$ \Rightarrow \,\sin \frac{\theta }{2} = \frac{{|a - b|}}{2}$.

 $\therefore a + b = 2i + 8k$=> $a - b = 2j$

Since, $(a + b)\,.\,(a - b) = 0$

 $\therefore (a + b) \bot (a - b)$. Hence $\theta = 90^\circ $.

$\sqrt {{x^2} + {y^2} + {z^2}} = 3 \Rightarrow {x^2} + {y^2} + {z^2} = 9$
.....$(i)$

$6x + 5y - 2z = 0$ .....$(ii)$

and $3x + y - 4z = 0$ .....$(iii)$

it is perpendicular to both vectors, hence by ${a_1}{b_1} + {a_2}{b_2} + {a_3}{b_3} = 0]$

On solving the equation $(i),$ $(ii)$ and $(iii),$ we get $x = 2,$ $y = - 2$ and $z = 1.$

Therefore, the required vector is $2i - 2j + k.$

Trick : By inspection, the vector $2i - 2j + k$ is of length $3 $ and also perpendicular to the given vectors.

Work done $ = F\,.\,d = - 4 + 6 - 4 = - 2$ or $2$ unit.

and displacement $d = 2i - 2j + 10k,$ then

Work done $W = F\,.\,d = 10 - 6 + 20 = 24\,unit.$

$ = (3i + 2j - 3k + 2i + 4j + 2k)\,.\,(5i + 4j + 2k - i - 2j - k)$

$ = (5i + 6j - k)\,.\,(4i + 2j + k)\, = 20 + 12 - 1 = 31$.

Force vector$ = 5\,.\,\left( {\frac{{2i - 2j + k}}{{|2i - 2j + k|}}} \right) = \frac{5}{3}(2i - 2j + k)$

$\therefore $ Required work done

$ = \frac{5}{3}(2i - 2j + k).[(5i + 3j + 7k) - (i + 2j + 3k)]$

$ = \frac{5}{3}[(2i - 2j + k).(4i + j + 4k)] = \frac{5}{3}[8 - 2 + 4] = \frac{{50}}{3}$unit.

Given, $|a + b{|^2} = {(\sqrt 3 )^2}$

==>$|a{|^2} + |b{|^2} + 2a.b = 3$

==> $2a.b = 1$ ==> $a.b = \frac{1}{2}$

Therefore, $(3a - 4b)\;.\;(2a + 5b)$

$ = 6 - 20 + 7 \times \frac{1}{2}$$ = - 14 + \frac{7}{2}$$ = \frac{{ - 21}}{2}$.

$ = j - k$

$\therefore $ Unit vector $ = \frac{{j - k}}{{\sqrt 2 }}$

$\left( {\because {\text{ }}\left( {\frac{{j - k}}{{\sqrt 2 }}} \right).(2i + j + k) = \frac{{ - 1 + 1}}{{\sqrt 2 }} = 0} \right)$

Then $M$ is the mid point of $AB.$

$\therefore $ Position vector of $M$ is $\frac{1}{2}(a + b).$

$\overrightarrow {PM} \,.\,\overrightarrow {BA} = 0$ or $\left[ {r - \frac{1}{2}(a + b)} \right]\,.\,(a - b) = 0.$

 $\therefore |\overrightarrow {AP} | = \sqrt {9 + 1 + 100} = \sqrt {110} $

$PN = $Projection of $\overrightarrow {AP} $ on $6i + 3j - 4k$

$ = \left| {\frac{{\overrightarrow {AP} .(6i + 3j - 4k)}}{{|6i + 3j - 4k|}}} \right|$$ = \left| {\frac{{ - 18 - 3 - 40}}{{\sqrt {61} }}} \right| = \sqrt {61} $

 $\therefore AN  = \sqrt {A{P^2} - A{N^2}} $$ = \sqrt {110 - 61} = 7$.

and $\overrightarrow {AC} = \overrightarrow {OC} - \overrightarrow {OA} = 2a + 3b - a = a + 3b.$

Let $\theta $ be the angle between $\overrightarrow {BD} $ and $\overrightarrow {AC} .$

Then $\cos \theta = \frac{{\overrightarrow {BD} \,.\,\overrightarrow {AC} }}{{|\overrightarrow {BD} |\,|\overrightarrow {AC} |}} = \frac{{|a{|^2} - 9|b{|^2}}}{{|\overrightarrow {BD} |\,|\overrightarrow {AC} |}}$

$ = \frac{{9|b{|^2} - 9|b{|^2}}}{{|\overrightarrow {BD} |\,|\overrightarrow {AC} |}}$,

$(\because \,\,\,\left| \,a\, \right|=3|b|)$

$ \Rightarrow \cos \theta = 0^\circ \Rightarrow \theta = \frac{\pi }{2}.$

$\Rightarrow$ $(\lambda + 1)^2 = 0$
$\Rightarrow$ $\lambda = -1$

$ \Rightarrow |\vec a{|^2} + |\vec b{|^2} - 2\left| {\vec a} \right|\left| {\vec b} \right|\cos \theta  = 1$

$ \Rightarrow  \cos \theta=1 / 2 \Rightarrow \cos \theta=1 / 2 $

$\Rightarrow  \theta=\pi / 3$

Then,  $|\vec{a}|=|\vec{b}|=1$

Now, $\vec{a}+\vec{b}$ is a unit vector if $|\vec{a}+\vec{b}|=1$

$|\vec{a}+\vec{b}|=1$

$\Rightarrow(\vec{a}+\vec{b})^{2}=1$

$\Rightarrow(\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})=1$

$\Rightarrow \vec{a} \cdot \vec{a}+\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{a}+\vec{b}\cdot \vec{b}=1$

$\Rightarrow|\vec{a}|^{2}+2 \vec{a} \cdot \vec{b}+|\vec{b}|^{2}=1$

$\Rightarrow 1^{2}+2|\vec{a}||\vec{b}| \cos \theta+1^{2}=1$

$\Rightarrow 1+2.1 .1 \cos \theta+1=1$

$\Rightarrow \cos \theta=-\frac{1}{2}$

$\Rightarrow \theta=\frac{2 \pi}{3}$

Hence, $\vec{a}+\vec{b}$ is a unit vector if $\theta=\frac{2 \pi}{3}$

The correct answer is $D$.

$\therefore \,\,\,a = (a\,.\,i)i + (a\,.\,j)j + (a\,.\,k)k$.

But $(i - j + k).({b_1}i + {b_2}j + {b_3}k) = 1$$ \Rightarrow {b_1} - {b_2} + {b_3} = 1$
......$(i)$

and $a \times b = \left| {\begin{array}{*{20}{c}}i&j&k\\1&{ - 1}&1\\{{b_1}}&{{b_2}}&{{b_3}}\end{array}} \right|$

$ = - i({b_2} + {b_3}) + j({b_1} - {b_3}) + k({b_2} + {b_1})$

$ \Rightarrow a \times b = c$

Comparing the coefficients of $i,\,\,j$ and $k$ respectively,

we get ${b_2} + {b_3} = 1$…..$(ii)$

${b_1} - {b_3} = - 1$…..$(iii)$

${b_2} + {b_1} = 0$…..$(iv)$

By solving the equations $(i),$ $(ii),$ $(iii)$ and $(iv),$ we get ${b_1} = 0,$ ${b_2} = 0$ and ${b_3} = 1$.

$ = (36 + 5n)\,i - (24 + 5m)\,j + (2n - 3m)\,k = 0$

$ \Rightarrow m = \frac{{ - 24}}{5},\,\,\,n = \frac{{ - 36}}{5}$.

$ = {a^2}{b^2}{\sin ^2}\theta = {a^2}{b^2}(1 - {\cos ^2}\theta )$

$ = {a^2}{b^2} - {a^2}{b^2}{\cos ^2}\theta = {a^2}{b^2} - {(a\,.\,b)^2}.$

$ \Rightarrow l{(a \times b)^2} = (c \times b)\,.\,(a \times b) \Rightarrow l = \frac{{(c \times b)\,.\,(a \times b)}}{{{{(a \times b)}^2}}}$

Similarly, $m = \frac{{(c \times a)\,.\,(b \times a)}}{{{{(b \times a)}^2}}}$.

$ \Rightarrow \,|j{|^2} + {(a\,.\,r)^2} = \,|a{|^2}|r{|^2}$$ \Rightarrow \,(a\,.\,r) = \pm \sqrt {|a{|^2}|r{|^2} - 1} $

This shows that $a\,.\,r$ depends on $|r|$ for given $a.$

Hence $a\,.\,r$ is arbitrary scalar.

and $i+j$  is $\frac{(i+j+k)\times (i+j)}{|(i+j+k)\times (i+j)|}$ 

, $ = \frac{{ - i + j}}{{\sqrt 2 }} = \frac{{ - 1}}{{\sqrt 2 }}(i - j)$ or where $ c$  is a scalar.

$ = \frac{1}{2}\left| {\begin{array}{*{20}{c}}i&j&k\\{ - 1}&0&1\\0&{ - 1}&1\end{array}} \right| = \frac{1}{2}(i + j + k)$

Hence by comparing, $\overrightarrow \alpha = i + j + k.$

$0$ , $a + b,\,\,a - b$ and $\theta = 90^\circ .$

Area of triangle $ = \frac{1}{2}|\overrightarrow {AB} \times \overrightarrow {AC} |$

$ = \frac{1}{2}|(a + b) \times (a - b)| = \frac{1}{2}|2b \times a|$

$ = ba\sin \theta = 3 \times 2\sin 90^\circ = 6.$

Then $p \times q = (2a - b) \times (4a - 5b) = - 6(a \times b)$

$ = - 6|a||b|\sin \frac{\pi }{4}\hat n = - 6 \times \frac{1}{{\sqrt 2 }}\hat n = - 3\sqrt 2 \,\hat n.$

Hence the area of the given parallelogram

$ = \frac{1}{2}|p \times q|\, = \frac{3}{{\sqrt 2 }}.$

$O\,(0,\,1,\,2)$ and $P\,(1,\, - 2,\,0) \Rightarrow \overrightarrow {OP} = i - 3j - 2k$

Resultant force $(F) = {F_1} + {F_2} + {F_3} = 4i + 4j + 2k$

Hence moment of force is $ = \overrightarrow {OP} \times F$

$ = \left| {\begin{array}{*{20}{c}}i&j&k\\1&{ - 3}&{ - 2}\\4&4&2\end{array}} \right| = 2i - 10j + 16k$

Magnitude of moment of force is

$|\overrightarrow {OP} \times F| = \sqrt {4 + 100 + 256} = 6\sqrt {10} .$

Moment of the force is $\overrightarrow {AP} \times \overrightarrow {AB} $
$ = \left| {\begin{array}{*{20}{c}}i&j&k\\3&0&{ - 3}\\4&4&{ - 1}\end{array}} \right| = 12i - 9j + 12k$

$\therefore $ Magnitude is, $\sqrt {144 + 81 + 144} = 3\sqrt {41} .$

$\therefore $ Moment $ = \overrightarrow {OA} \times F = \frac{6}{{11}}\left| {\,\begin{array}{*{20}{c}}i&j&k\\3&2&{ - 9}\\9&6&{ - 2}\end{array}\,} \right|$

$ = \frac{6}{{11}}(50i - 75j) = \frac{{150}}{{11}}(2i - 3j)$.

$n = a \times b$; $n = $ $\frac{{\left| {\begin{array}{*{20}{c}}i&j&k\\1&{ - 1}&0\\1&1&0\end{array}} \right|}}{{\sqrt 2 \times \sqrt 2 }}$ = $\frac{{2k}}{2} = k$

 $\therefore |c\;.\;n|\; = \,|(i + 3j + 5k).(k)|\, = \;|5|\, = 5$.

$\therefore $ Required vector is $ \pm \frac{{(i - j + k) \times ( - i + j + k)}}{{\left| {(i - j + k) \times ( - i + j + k)} \right|}}$

$ = \pm \frac{{( - (i + j))}}{{\sqrt 2 }}i.e.,\;\frac{{ - (i + j)}}{{\sqrt 2 }}$ and $\frac{{i + j}}{{\sqrt 2 }}$.

$ = 2i - 6j + 5k$

Moment of Force $\overrightarrow F $ w.r.t $M = \overrightarrow {MA} \times \overrightarrow F $

$\because \,\,\,\overrightarrow {MA}  = \left( {1 + 2} \right)\,i + \left( {2 - 4} \right)\,j + \left( { - 3 + 6} \right)\,k = 3i - 2j + 3k$

Now $\overrightarrow {MA} \times \overrightarrow F = \left| {\,\begin{array}{*{20}{c}}i&j&k\\3&{ - 2}&3\\2&{ - 6}&5\end{array}\,} \right|$

$ = i\,\left( { - 10 + 18} \right) + j\,\left( {6 - 15} \right) + k\,\left( { - 18 + 4} \right)$

$ = 8i - 9j - 14k$.

Adding $r \times (a + b) = 0$i.e., r is parallel to $a + b$

or $r = \lambda (i + j + 2i - k)$

$r = \lambda (3i + j - k)$ for $\lambda = 1 \Rightarrow r = (3i + j - k)$.

 

We know that $\vec{a} \times \vec{b}=|\vec{a}||\vec{b}| \sin \theta \,\hat{n},$ where is a unit vector perpendicular to both $\vec{a}$ and $\vec{b}$ and $\theta$ is the angle between $\vec{a}$ and $\vec{b}$.

Now, $\vec{a} \times \vec{b}$ is a unit vector if $|\vec{a} \times \vec{b}|=1$

$|\vec{a} \times \vec{b}|=1$

$\Rightarrow|| \vec{a}|| \vec{b}|\sin \theta \hat{n}|=1$

$\Rightarrow|\vec{a}||\overrightarrow{ b }||\sin \theta|=1$

$\Rightarrow 3 \times \frac{\sqrt{2}}{3} \times \sin \theta=1$

$\Rightarrow \sin \theta=\frac{1}{\sqrt{2}}$

$\Rightarrow \theta=\frac{\pi}{4}$

Hence, $\vec{a} \times \vec{b}$ is a unit vector if the angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{4}$

The correct answer is $A$.

$ = \lambda \,[a\,b\,c] + 0 + 0 = \lambda \,[a\,b\,c] = \frac{\lambda }{8}$

Hence $\lambda = 8(d\,.\,c),$ $\mu = 8(d\,.\,a)$ and $\nu = 8(d\,.\,b)$

Therefore, $\lambda + \mu + \nu = 8d\,.\,c + 8d.\,a + 8d\,.\,b$

$ = 8d\,.\,(a + b + c).$

$(a + b + c)\,.\,(p + q + r) = \frac{{[a\,b\,c] + [b\,c\,a] + [c\,a\,b]}}{{[a\,b\,c]}} = 3$.

$ \Rightarrow {(ab + bc + ca)^3} = 0 \Rightarrow ab + bc + ca = 0.$

$=\frac{(b\times c)\,.\,a+0+0}{\lambda }$
$ = \frac{\lambda }{\lambda } = 1$,

($\because $ Given $a.\,(b\times c)=\lambda =(b\times c)\,.\,a$)

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

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

$[\because \,\,b \times b = 0\,o\,c \times c = 0]$

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

$ = - \,[a\,a\,b] + [a\,c\,a] - [b\,a\,b] + [b\,c\,a]$

$ = 0 + 0 - 0 + [b\,c\,a]$$ = [a\,b\,c]$.

So, $\overrightarrow {AB} = - i - j + 4k,\,\,\overrightarrow {BC} = 2i + 2j - 5k,$

$\overrightarrow {CD} = - 2i - (\lambda + 4)j + 3k$

From the condition, $[\overrightarrow {AB} \,\,\overrightarrow {BC} \,\,\overrightarrow {CD} ] = 0$

==> $\left| {\,\begin{array}{*{20}{c}}{ - 1}&{ - 1}&4\\2&2&{ - 5}\\{ - 2}&{ - (\lambda + 4)}&3\end{array}\,} \right| = 0$

==> $ - 1[2.3 - 5(\lambda + 4)] + 1[6 - 10] + 4[ - 2(\lambda + 4) + 4] = 0$

$ \Rightarrow \lambda = - 2$.