Question 13 Marks
For any two vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ of magnitudes 3 and 4 respectively, write the value of $\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{a}}\times\vec{\text{b}}\big]+\big(\vec{\text{a}}.\vec{\text{b}}\big)^2$
AnswerWe have
$\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{a}}\times\vec{\text{b}}\big]+\big(\vec{\text{a}}.\vec{\text{b}}\big)^2$
$=\big[\big(\vec{\text{a}}\times\vec{\text{b}}\big).\big(\vec{\text{a}}\times\vec{\text{b}}\big)\big]+\big(\vec{\text{a}}.\vec{\text{b}}\big)^2$ (By definition of scalar triple product)
$=\big|\big(\vec{\text{a}}\times\vec{\text{b}}\big)\big|^2+\big(\vec{\text{a}}.\vec{\text{b}}\big)^2$
$=\big(\big|\vec{\text{a}}\big|\big|\vec{\text{b}}\big|\sin\theta\big)^2+\big(\big|\vec{\text{a}}\big|\big|\vec{\text{b}}\big|\cos\theta\big)^2$
$=\big(\big|\vec{\text{a}}\big|\big|\vec{\text{b}}\big|\big)^2\sin^2\theta+\big(\big|\vec{\text{a}}\big|\big|\vec{\text{b}}\big|\big)^2\cos^2\theta$
$=\big(\big|\vec{\text{a}}\big|\big|\vec{\text{b}}\big|\big)^2(\sin^2\theta+\cos^2\theta)$
$=\big(\big|\vec{\text{a}}\big|\big|\vec{\text{b}}\big|\big)^2$ $\big(\therefore\sin^2\theta+\cos^2\theta=1\big)$
$=(3\times4)^2$ $\text{(Given}:\big|\vec{\text{a}}\big|=3\text{ and } \big|\vec{\text{b}}\big|=4\big)$ $$
$=144$
View full question & answer→Question 23 Marks
Find the volume of the parallelopiped whose coterminous edges are represented by the vectors:
$\vec{\text{a}}=11\hat{\text{i}},\vec{\text{b}}=2\hat{\text{j}},\vec{\text{c}}=13\hat{\text{k}}$
AnswerGiven,
$\vec{\text{a}}=11\hat{\text{i}}$
$\vec{\text{b}}=2\hat{\text{j}}$
$\vec{\text{c}}=13\hat{\text{k}}$
We know that the volume of a parallelopiped whose three adjacent edges are $\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}$
is equal to $\Big|\Big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\Big]\Big|$ Here,
$\Big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\Big]=\begin{vmatrix}11&0&0\\0&2&0\\0&0&13 \end{vmatrix}$
$=11(26-0)-0(0-0)+0(0-0)$
$=286$
volume of the parallelopiped $=\Big|\Big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\Big]\Big|=|286|=286$ cubic units.
View full question & answer→Question 33 Marks
Find the volume of the parallelopiped whose coterminous edges are represented by the vectors:
$\vec{\text{a}}=2\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}},\vec{\text{b}}=\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}},\vec{\text{c}}=3\hat{\text{i}}-\hat{\text{j}}-2\hat{\text{k}}$
AnswerGiven,
$\vec{\text{a}}=2\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}}$
$\vec{\text{b}}=\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}$
$\vec{\text{c}}=3\hat{\text{i}}-\hat{\text{j}}-2\hat{\text{k}}$
We know that the volume of a parallelopiped whose three adjacent edges are $\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}$ Is equal to $\Big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\Big]$ Here,
$\Big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\Big]=\begin{vmatrix}2&-3&4\\1&2&-1\\3&-1&-2 \end{vmatrix}$
$=2(-4-1)+3(2+3)+4(-1-6)$
$=-35$
volume of the parallelopiped $=\Big|\Big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\Big]\Big|=|-35|=35$ cubic units.
View full question & answer→Question 43 Marks
Let $\vec{\text{a}}=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}},\vec{\text{b}}=\hat{\text{i}}$ and $\vec{\text{c}}=\text{c}_1\hat{\text{i}}+\text{c}_2\hat{\text{j}}+\text{c}_3\hat{\text{k}}.$ Then,
If $c_1= 1$ and $c_2= 2$, find $c_3$which makes $\vec{\text{a}},\vec{\text{b}}$ and $\vec{\text{c}}$ coplanar.
AnswerIf $c_1 = 1$ and $c_2= 2$, then $\vec{\text{a}}=\hat{\text{i}}+\vec{\text{j}}+\vec{\text{k}},\vec{\text{b}}=\hat{\text{i}}$ and $\vec{\text{c}}=\hat{\text{i}}+2\hat{\text{j}}+\text{c}_3\hat{\text{k}}.$
We know that vectors $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are coplanar if $\big[\vec{\text{a}}\vec{\text{b}}{\text{c}}\big]=0.$
It is given that $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are coplanar.
$\therefore\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]=0$
$\Rightarrow\begin{vmatrix}1&1&1\\1&0&0\\1&2&\text{c}_3 \end{vmatrix}=0$
$\Rightarrow1(0-0)-1(\text{c}_3-0)+1(2-0)=0$
$\Rightarrow-{\text{C}}_3+2=0$
$\Rightarrow\text{C}_3=2$
View full question & answer→Question 53 Marks
Find the volume of the parallelopiped whose coterminous edges are represented by the vectore:
$\vec{\text{a}}=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}},\vec{\text{b}}=\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}},\vec{\text{c}}=\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}$
Answer$\vec{\text{a}}=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$
$\vec{\text{b}}=\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$
$\vec{\text{c}}=\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}$
We know that the volume of a parallelopiped whose three adjacent edges
are $\vec{\text{a}},\ \vec{\text{b}},\ \vec{\text{c}}$ is equal to $\Big|\Big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\Big]\Big|$ Here,
$\Big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\Big]=\begin{vmatrix}1&1&1\\1&-1&1\\1&2&-1 \end{vmatrix}\\=1(1-2)-1(-1-1)+1(2+1)=4$
Volume of the parallelopiped $=\Big[\Big|\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\Big]\Big|=|4|=4$ cubic units.
View full question & answer→Question 63 Marks
Find the volume of the parallelopiped whose coterminous edges are represented by the vectors:
$\vec{\text{a}}=2\hat{\text{i}}+3\hat{\text{j}}+4\hat{\text{k}},\vec{\text{b}}=\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}},\vec{\text{c}}=3\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}$
AnswerGiven:
$\vec{\text{a}}=2\hat{\text{i}}+3\hat{\text{j}}+4\hat{\text{k}}$
$\vec{\text{b}}=\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}$
$\vec{\text{c}}=3\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}$
We know that the volume of a parallelopiped whose three adjacent edges are $\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}$ Is equal to $\Big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\Big]$ Here,
$\Big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\Big]=\begin{vmatrix}2&3&4\\1&2&-1\\3&-1&2 \end{vmatrix}$
$=2(4-1)-3(2+3)+4(-1-6)=-37$
volume of the parallelopiped $=\Big|\Big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\Big]\Big|=|-37|=37$ cubic units.
View full question & answer→Question 73 Marks
Show that the four points having position vectors
$6\hat{\text{i}}-7\hat{\text{j}},16\hat{\text{i}}-19\hat{\text{j}}-4\hat{\text{k}},3\hat{\text{j}}-6\hat{\text{k}},2\hat{\text{i}}+5\hat{\text{j}}+10\hat{\text{k}}$ are not coplanar.
AnswerLet
$\text{OA}=6\hat{\text{i}}-7\hat{\text{j}},{\text{OB}}=16\hat{\text{i}}-19\hat{\text{j}}-4\hat{\text{k}},$
$\text{OC}=3\hat{\text{j}}-6\hat{\text{k}},\text{OD}=2\hat{\text{i}}+5\hat{\text{j}}+10\hat{\text{k}}$
$\text{AB}=\text{OB}-\text{OA}=16\hat{\text{i}}-25\hat{\text{j}}-4\hat{\text{k}}$
$\text{AC}=\text{OC}-\text{OA}=-16\hat{\text{i}}-16\hat{\text{j}}+2\hat{\text{k}}$
$\text{CD}=\text{OD}-\text{OC}=2\hat{\text{i}}+2\hat{\text{j}}+16\hat{\text{k}}$
$\text{AD}=\text{OD}-\text{OA}=4\hat{\text{i}}+12\hat{\text{j}}+10\hat{\text{k}}$
The four points are co-planer if vectors $\vec{\text{AB}},\vec{\text{AC}},\vec{\text{AD}}$ are co-planer.
$\begin{vmatrix}16&-25&-4\\-16&-16&2\\-4&12&10 \end{vmatrix}$
$=16(-160-24)+25(-160+8)-4(-144+64)$
$\neq 0$
Hence the point are not co-planar.
View full question & answer→Question 83 Marks
Let $\vec{\text{a}}=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}},\vec{\text{b}}=\hat{\text{i}}$ and $\vec{\text{c}}=\text{c}_1\hat{\text{i}}+\text{c}_2\hat{\text{j}}+\text{c}_3\hat{\text{k}}.$ Then,
If $C_2 = -1$ and $C_3 = 1$, show that no value of $C_1$ can make $\vec{\text{a}},\vec{\text{b}}$ and $\vec{\text{c}}$ coplanar.
AnswerIf $C_2= -1$ and $C_3 = 1$, then $\vec{\text{a}}=\hat{\text{i}}+\vec{\text{j}}+\vec{\text{k}},\vec{\text{b}}=\hat{\text{i}}$ and $\vec{\text{c}}={\text{c}}_1\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}.$
We know that vectors $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are coplanar if $\big[\vec{\text{a}}\vec{\text{b}}{\text{c}}\big]=0.$
for $\big[\vec{\text{a}},\vec{\text{b}},\text{c}\big]$ to be coplanar:
$\therefore\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]=0$
$\Rightarrow\begin{vmatrix}1&1&1\\1&0&0\\\text{c}_1&-1&1 \end{vmatrix}=0$
$\Rightarrow 1(0-0)-1(1-0)+1(-1-0)=0$
$\Rightarrow-1-1=0$
$\Rightarrow -2=0$
But this is never possible, whatever be the value of $C_1$_. Thus, no value of $C_1$_ can make $\vec{\text{a}},\vec{\text{b}}$ and $\vec{\text{c}}$ coplanar.
View full question & answer→Question 93 Marks
Find $\big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\big]$, when
$\vec{\text{a}}=2\hat{\text{i}}-3\hat{\text{j}},\vec{\text{b}}=\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}$ and $\vec{\text{c}}=3\hat{\text{i}}-\hat{\text{k}}$
AnswerGiven:
$\vec{\text{a}}=2\hat{\text{i}}-3\hat{\text{j}}$
$\vec{\text{b}}=\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}$
$\vec{\text{c}}=3\hat{\text{i}}-\hat{\text{k}}$
$\therefore\vec{\text{a}}\times\vec{\text{b}}=(2\hat{\text{i}}-3\hat{\text{j}})\times(\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}})$
$=2\hat{\text{k}}+2\hat{\text{j}}+3\hat{\text{k}}+3\hat{\text{i}}$
$=3\hat{\text{i}}+2\hat{\text{j}}+5\hat{\text{k}}$
$\big(\vec{\text{a}}\times\vec{\text{b}}\big).\vec{\text{c}}=\big(3\hat{\text{i}}+2\hat{\text{j}}+5\hat{\text{k}}\big).\big(3\hat{\text{i}}-\hat{\text{k}}\big)$
$=9-5=4\ ....(1)$
Now,
$\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]=\big(\vec{\text{a}}\times\vec{\text{b}}\big).\vec{\text{c}}$
$=4$ [Using (1)]
View full question & answer→Question 103 Marks
If $\big[3\vec{\text{a}}+7\vec{\text{b}}\vec{\text{c}}\vec{\text{d}}\big]=\lambda\big[\vec{\text{a}}\vec{\text{c}}\vec{\text{d}}\big]+\mu\big[\vec{\text{b}}\vec{\text{c}}\vec{\text{d}}\big],$ then find the value of $\lambda+\mu.$
AnswerWe have
$\big[3\vec{\text{a}}+7\vec{\text{b}}\vec{\text{c}}\vec{\text{d}}\big]=\lambda\big[\vec{\text{a}}\vec{\text{c}}\vec{\text{d}}\big]+\mu\big[\vec{\text{b}}\vec{\text{c}}\vec{\text{d}}\big]$
$\Rightarrow\big[\big(3\vec{\text{a}}+7\vec{\text{b}}\big)\times\vec{\text{c}}\big].\vec{\text{d}}=\lambda\big[\vec{\text{a}}\vec{\text{c}}\vec{\text{d}}\big]+\mu\big[\vec{\text{b}}\vec{\text{c}}\vec{\text{d}}\big]$(By definition of scalar triple product)
$\Rightarrow\big[\big(3\vec{\text{a}}\times\vec{\text{c}}\big)+\big(7\vec{\text{b}}\times\vec{\text{c}}\big)\big].\vec{\text{d}}=\lambda\big[\vec{\text{a}}\vec{\text{c}}\vec{\text{d}}\big]+\mu\big[\vec{\text{b}}\vec{\text{c}}\vec{\text{d}}\big]$
$\Rightarrow\big(3\vec{\text{a}}\times\vec{\text{c}}\big].\vec{\text{d}}+\big(7\vec{\text{b}}\times\vec{\text{c}}\big).\vec{\text{d}}=\lambda\big[\vec{\text{a}}\vec{\text{c}}\vec{\text{d}}\big]+\mu\big[\vec{\text{b}}\vec{\text{c}}\vec{\text{d}}\big]$
$\Rightarrow\big[3\vec{\text{a}}\vec{\text{c}}\vec{\text{d}}\big]+\big[7\vec{\text{b}}\vec{\text{c}}\vec{\text{d}}\big]=\lambda\big[\vec{\text{a}}\vec{\text{c}}\vec{\text{d}}\big]+\mu\big[\vec{\text{b}}\vec{\text{c}}\vec{\text{d}}\big]$
$\Rightarrow3\big[\vec{\text{a}}\vec{\text{c}}\vec{\text{d}}\big]+7\big[\vec{\text{b}}\vec{\text{c}}\vec{\text{d}}\big]=\lambda\big[\vec{\text{a}}\vec{\text{c}}\vec{\text{d}}\big]+\mu\big[\vec{\text{b}}\vec{\text{c}}\vec{\text{d}}\big]$
$\big(\therefore\big[\lambda\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]=\lambda\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]$ for any scalar $\lambda\big)$
Comparing both sides, we get
$\lambda=3$
$\mu=7$
$\therefore\lambda+\mu=3+7=10$
View full question & answer→Question 113 Marks
Find the value of $\lambda$ so that the following vectors are coplanar:
$\vec{\text{a}}=\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}},\vec{\text{b}}=2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}},\vec{\text{c}}=\lambda\hat{\text{i}}-\hat{\text{j}}+\lambda\hat{\text{k}}$
AnswerGiven:
$\vec{\text{a}}=\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$
$\vec{\text{b}}=2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}$
$\vec{\text{c}}=\lambda\hat{\text{i}}-\hat{\text{j}}+\lambda\hat{\text{k}}$
We know that vectors $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are coplanar if $\Big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\Big]=0$
It is given that $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are coplanar.
$\therefore\Big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\Big]=0$
$\Rightarrow\begin{vmatrix}1&-1&1\\2&1&-1\\\lambda&-1&\lambda \end{vmatrix}=0$
$\Rightarrow 1(\lambda-1)+1(2\lambda+\lambda)+1(-2-\lambda)=0$
$\Rightarrow \lambda -1+3\lambda-2-\lambda=0$
$\Rightarrow 3\lambda-3=0$
$\Rightarrow\lambda=1$
View full question & answer→Question 123 Marks
Find $\Big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\Big]$, when
$\vec{\text{a}}=\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}},\vec{\text{b}}=2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}$ and $\vec{\text{c}}=\hat{\text{j}}+\hat{\text{k}}$
AnswerGiven:
$\vec{\text{a}}=\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}$
$\vec{\text{b}}=2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}$
$\vec{\text{c}}=\hat{\text{j}}+\hat{\text{k}}$
$\therefore\vec{\text{a}}\times\vec{\text{b}}=\big(\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}\big)\times\big(2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}\big)$
$=\hat{\text{k}}+\hat{\text{j}}+4\hat{\text{k}}+2\hat{\text{i}}+6\hat{\text{j}}-3\hat{\text{i}}$
$=-\hat{\text{i}}+7\hat{\text{j}}+5\hat{\text{k}}$
$\big(\vec{\text{a}}\times\vec{\text{b}}\big).\vec{\text{c}}=\big(-\hat{\text{i}}+7\hat{\text{j}}+5\hat{\text{k}}\big).\big(\hat{\text{j}}+\hat{\text{k}}\big)$
$=7+5=12\ ...(1)$
Now,
$\big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\big]=\big(\vec{\text{a}}\times\vec{\text{b}}\big).\vec{\text{c}}$
$=12$ [Using (1)]
View full question & answer→Question 133 Marks
Find the value of $\lambda$ so that the following vectors are coplanar:
$\vec{\text{a}}=\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}},\vec{\text{b}}=3\hat{\text{i}}+\lambda\hat{\text{j}}+\hat{\text{k}},\vec{\text{c}}=\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}$
AnswerGiven:
$\vec{\text{a}}=\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}$
$\vec{\text{b}}=3\hat{\text{i}}+\lambda\hat{\text{j}}+\hat{\text{k}}$
$\vec{\text{c}}=\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}$
We know that vectors $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are coplanar iff $\Big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\Big]=0.$
It is given that $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are coplanar.
$\therefore\Big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\Big]=0$
$\Rightarrow\begin{vmatrix}1&2&-3\\3&\lambda&1\\1&2&2 \end{vmatrix}=0$
$\Rightarrow1(2\lambda-2)-2(6-1)-3(6-\lambda)=0$
$\Rightarrow5\lambda-30=0$
$\Rightarrow\lambda=6$
View full question & answer→Question 143 Marks
Find the value of $\lambda$ so that the following vectors are coplanar:
$\vec{\text{a}}=2\hat{\text{i}}-\hat{\text{j}}++\hat{\text{k}},\vec{\text{b}}=\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}},\vec{\text{c}}=\lambda\hat{\text{i}}+\lambda\hat{\text{j}}+5\hat{\text{k}}$
AnswerGiven:
$\vec{\text{a}}=2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$
$\vec{\text{b}}=\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}$
$\vec{\text{c}}=\lambda\hat{\text{i}}+\lambda\hat{\text{j}}+5\hat{\text{k}}$
We know that vectors $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are coplanar iff $\Big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\Big]=0.$
It is given that $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are coplanar.
$\therefore\Big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\Big]=0$
$\Rightarrow\begin{vmatrix}2&-1&1\\1&2&-3\\\lambda&\lambda&5 \end{vmatrix}=0$
$\Rightarrow2(10+3\lambda)+1(5+3\lambda)+1(\lambda-2\lambda)=0$
$\Rightarrow 8\lambda +25=0$
$=-\frac{25}{8}$
View full question & answer→