Question 13 Marks
If $\vec{\text{a}},\vec{\text{b}}$ are two vectors such that $\big|\vec{\text{a}}+\vec{\text{b}}\big|=\big|\vec{\text{b}}\big|,$ then prove that $\vec{\text{a}}+2\vec{\text{b}}$ is perpendicular to $\vec{\text{a}}.$
AnswerGiven that
$\big|\vec{\text{a}}+\vec{\text{b}}\big|=\big|\vec{\text{b}}\big|$
Squaring both sides, we get
$\big|\vec{\text{a}}+\vec{\text{b}}\big|^2=\big|\vec{\text{b}}\big|^2$
$\Rightarrow|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2+2\vec{\text{a}}.\vec{\text{b}}=\big|\vec{\text{b}}\big|^2$
$\Rightarrow|\vec{\text{a}}|^2+2\vec{\text{a}}.\vec{\text{b}}=0\dots(1)$
Now,
$\big(\vec{\text{a}}+2\vec{\text{b}}\big).\vec{\text{a}}$
$\vec{\text{a}}.\vec{\text{a}}+2\vec{\text{b}}.\vec{\text{a}}$
$=|\vec{\text{a}}|^2+2\vec{\text{a}}.\vec{\text{b}}$
$=0$ [Using (1)]
So, $\vec{\text{a}}+2\vec{\text{b}}$ is perpendicular to $\vec{\text{a}}.$
View full question & answer→Question 23 Marks
If $\vec{\text{a}}=2\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}},\vec{\text{b}}=-\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}},$and $\vec{\text{c}}=3\hat{\text{i}}+\hat{\text{j}}$ are such that $\vec{\text{a}}+\lambda\vec{\text{b}}$ is perpendicular to $\vec{\text{c}},$ then find the value of $\lambda.$
AnswerThe given vectors are $\vec{\text{a}}=2\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}},\vec{\text{b}}=-\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}},$ and $\vec{\text{c}}=3\hat{\text{i}}+\hat{\text{j}}$.
Now,
$\vec{\text{a}}+\lambda\vec{\text{b}}=\big(2\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\big)+\lambda\big(-\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}\big)\\=(2-\lambda)\hat{\text{i}}+(2+2\lambda)\hat{\text{j}}+(3+\lambda)\hat{\text{k}}$
If $\big(\vec{\text{a}}+\lambda\vec{\text{b}}\big)$ is perpendicular to $\vec{\text{c}},$ then
$\big(\vec{\text{a}}+\lambda\vec{\text{b}}\big).\vec{\text{c}}=0.$
$\Rightarrow\Big[\big(2-\lambda\big)\hat{\text{i}}+(2+2\lambda)\hat{\text{j}}+(3+\lambda)\hat{\text{k}}\Big].\big(3\hat{\text{i}}+\hat{\text{j}}\big)=0$
$\Rightarrow(2-\lambda)3+(2+2\lambda)1+(3+\lambda)0=0$
$\Rightarrow6-3\lambda+2+2\lambda=0$
$\Rightarrow-\lambda+8=0$
$\Rightarrow\lambda=8$
Hence, the required value of $\lambda$ is 8.
View full question & answer→Question 33 Marks
Find $\lambda$ when the projection of $\vec{\text{a}}=\lambda\hat{\text{i}}+\hat{\text{j}}+4\hat{\text{k}}$ on $\vec{\text{b}}=2\hat{\text{i}}+6\hat{\text{j}}+3\hat{\text{k}}$ is 4 units.
AnswerWe know
$\vec{\text{a}}=\lambda\hat{\text{i}}+\hat{\text{j}}+4\hat{\text{k}}$ and $\vec{\text{b}}=2\hat{\text{i}}+6\hat{\text{j}}+3\hat{\text{k}}$
The projection of $\vec{\text{a}}$ on $\vec{\text{b}}$ is $\frac{\vec{\text{a}}.\vec{\text{b}}}{\big|\vec{\text{b}}\big|}$
Given that
$\frac{\vec{\text{a}}.\vec{\text{b}}}{\big|\vec{\text{b}}\big|}=4$
$\Rightarrow\frac{\big(\lambda\hat{\text{i}}+\hat{\text{j}}+4\hat{\text{k}}\big).\big(2\hat{\text{i}}+6\hat{\text{j}}+3\hat{\text{k}}\big)}{\big|2\hat{\text{i}}+6\hat{\text{j}}+3\hat{\text{k}}\big|}$
$\Rightarrow\frac{2\lambda+6+12}{\sqrt{4+36+9}}=4$
$\Rightarrow\frac{2\lambda+18}{7}=4$
$\Rightarrow2\lambda+18=28$
$\Rightarrow2\lambda=10$
$\therefore\lambda=5$
View full question & answer→Question 43 Marks
Dot product of a vector with vectore $\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}},2\hat{\text{i}}+\hat{\text{j}}-3\hat{\text{k}}$ and $\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$ are respectively 4, 0 and 2. Find the vector.
AnswerLet $\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}}$ be the required vector.
Given that
$\big(\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}}\big).\big(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}\big)=4$
$\Rightarrow\text{a}-\text{b+c}=4\dots(1)$
$\big(\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}}\big).\big(2\hat{\text{i}}+\hat{\text{j}}-3\hat{\text{k}}\big)=0$
$\Rightarrow2\text{a}+\text{b}-3\text{c}=0\dots(2)$
$\big(\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}}\big).\big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)=2$
$\Rightarrow\text{a+b}+\text{c}=2\dots(3)$
Solving (1), (2) and (3), we get
$\text{a}=2,\text{b}=-1,\text{c}=1$
So, $\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}}=2\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$
View full question & answer→Question 53 Marks
Let $\vec{\text{a}}=5\hat{\text{i}}-\hat{\text{j}}+7\hat{\text{k}}$ and $\vec{\text{b}}=\hat{\text{i}}-\hat{\text{j}}+\lambda\hat{\text{k}}.$ Find $\lambda$ such that $\vec{\text{a}}+\vec{\text{b}}$ is orthonal to $\vec{\text{a}}-\vec{\text{b}}.$
AnswerGive that
$\vec{\text{a}}=5\hat{\text{i}}-\hat{\text{j}}+7\hat{\text{k}};\vec{\text{b}}=\hat{\text{i}}-\hat{\text{j}}+\lambda\hat{\text{k}}$
$\therefore\vec{\text{a}}+\vec{\text{b}}=5\hat{\text{i}}-\hat{\text{j}}+7\hat{\text{k}}+\hat{\text{i}}-\hat{\text{j}}+\lambda\hat{\text{k}}=6\hat{\text{i}}-2\hat{\text{j}}+(7+\lambda)\hat{\text{k}}$
and $\vec{\text{a}}-\vec{\text{b}}=5\hat{\text{i}}-\hat{\text{j}}+7\hat{\text{k}}-\big(\hat{\text{i}}-\hat{\text{j}}+\lambda\hat{\text{k}}\big)=4\hat{\text{i}}+0\hat{\text{j}}+(7-\lambda)\hat{\text{k}}$
Given that $\vec{\text{a}}+\vec{\text{b}}$ is orthogonal to $\vec{\text{a}}-\vec{\text{b}}.$
$\Rightarrow\big(\vec{\text{a}}+\vec{\text{b}}\big).\big(\vec{\text{a}}-\vec{\text{b}}\big)=0$
$\Rightarrow\big[6\hat{\text{i}}-2\hat{\text{j}}+(7+\lambda)\hat{\text{k}}\big].\big[4\hat{\text{i}}+0\hat{\text{j}}+(7-\lambda)\hat{\text{k}}\big]=0$
$\Rightarrow24+0+49-\lambda^2=0$
$\Rightarrow\lambda^2=73$
$\Rightarrow\lambda=\sqrt{73}$
View full question & answer→Question 63 Marks
If $\vec{\text{a}}$ and $\vec{\text{b}}$ are perpendicular vectors, $\big|\vec{\text{a}}+\vec{\text{b}}\big|=3$ and $|\vec{\text{a}}|=5,$ find the value of $\big|\vec{\text{b}}\big|.$
AnswerDisclamer: $\big|\vec{\text{a}}+\vec{\text{b}}\big|=3$ has been taken in order to solve question.
It is given that $\vec{\text{a}}$ and $\vec{\text{b}}$ are perpendicular vectors.
$\therefore\vec{\text{a}}.\vec{\text{b}}=0\dots(1)$
$\big|\vec{\text{a}}+\vec{\text{b}}\big|=13$
$\Rightarrow\big|\vec{\text{a}}+\vec{\text{b}}\big|^2=169$
$\Rightarrow|\vec{\text{a}}|^2+2\vec{\text{a}}.\vec{\text{b}}+\big|\vec{\text{b}}\big|^2=169$
$\Rightarrow25+2\times0+\big|\vec{\text{b}}\big|^2=169$ [using (1)]
$\Rightarrow\big|\vec{\text{b}}\big|^2=169-25=144$
$\Rightarrow\big|\vec{\text{b}}\big|=12$
Thus, the value of $\big|\vec{\text{b}}\big|$ is 12.
View full question & answer→Question 73 Marks
If a vector$\vec{\text{a}}$ is perpendicular to two non-collinear vectors $\vec{\text{b}}$ and $\vec{\text{c}},$ then show that $\vec{\text{a}}$ is perpendicular to every vector in the plane of $\vec{\text{b}}$ and $\vec{\text{c}}.$
AnswerGiven that $\vec{\text{a}}$ is perpendicular to $\vec{\text{b}}$ and $\vec{\text{c}}.$
$\Rightarrow\vec{\text{a}}.\vec{\text{b}}=0$ and $\vec{\text{a}}.\vec{\text{c}}=0\dots(1)$
Now, let $\vec{\text{r}}$ be any vector in the plane of $\vec{\text{b}}$ and $\vec{\text{c}}.$
Then, $\vec{\text{r}}$ is the linear combination of $\vec{\text{b}}$ and $\vec{\text{c}}.$
$\vec{\text{r}}=\text{x}\vec{\text{b}}+\text{y}\vec{\text{c}}, $ for some x and y.
Now,
$\vec{\text{a}}.\vec{\text{r}}$
$=\vec{\text{a}}.\big(\text{x}\vec{\text{b}}+\text{y}\vec{\text{c}}\big)$
$=\text{x}\big(\vec{\text{a}}.\vec{\text{b}}\big)+\text{y}\big(\vec{\text{a}}.\vec{\text{c}}\big)$
$=\text{x}(0)+\text{y}(0)$ [From(1)]
$=0$
Thus, $\vec{\text{a}}$ is perpendicular to $\vec{\text{r}}.$
That is, $\vec{\text{a}}$ is perpendicular to every vector in the plane of $\vec{\text{b}}$ and $\vec{\text{c}}.$
View full question & answer→Question 83 Marks
If $\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}=\vec{0,}$ show that the angle $\theta$ between the vectors $\vec{\text{b}}$ and $\vec{\text{c}}$ is given by $\cos\theta=\frac{|\vec{\text{a}}|^2-\big|\vec{\text{b}}\big|^2-|\vec{\text{c}}|^2}{2\big|\vec{\text{b}}\big||\vec{\text{c}}|}.$
AnswerGiven,$\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}=0$
$\Rightarrow\vec{\text{b}}+\vec{\text{c}}=-\vec{\text{a}}$
$\Rightarrow\big|\vec{\text{b}}+\vec{\text{c}}\big|^2=|-\vec{\text{a}}|^2$
$\Rightarrow\big|\vec{\text{b}}\big|^2+|\vec{\text{c}}|^2+2\vec{\text{b}}.\vec{{\text{c}}}=|\vec{\text{a}}|^2$
$\Rightarrow2\vec{\text{b}}.\vec{\text{c}}=|\vec{\text{a}}|^2-\big|\vec{\text{b}}\big|^2-|\vec{\text{c}}|^2$
$\Rightarrow2\big|\vec{\text{b}}\big||\vec{\text{c}}|\cos\theta=|\vec{\text{a}}|^2-\big|\vec{\text{b}}\big|^2-|\vec{\text{c}}|^2$
$\therefore\cos\theta=\frac{|\vec{\text{a}}|^2-\big|\vec{\text{b}}\big|^2-|\vec{\text{c}}|^2}{2\big|\vec{\text{b}}\big||\vec{\text{c}}|}$
View full question & answer→Question 93 Marks
Find $|\vec{\text{a}|}$ and $\big|\vec{\text{b}}\big|$ if
$\big(\vec{\text{a}}+\vec{\text{b}}\big).\big(\vec{\text{a}}-\vec{\text{b}}\big)=8$ and $|\vec{\text{a}}|=8\big|\vec{\text{b}}\big|$
Answer$\big(\vec{\text{a }}.\vec{\text{b}}\big).\big(\vec{\text{a}}-\vec{\text{b}}\big)=8$
$\Rightarrow \vec{\text{a}}.\vec{\text{a}}-\vec{\text{a}}.\vec{\text{b}}+\vec{\text{b}}.\vec{\text{a}}-\vec{\text{b}}.\vec{\text{b}}=8$
$\Rightarrow|\vec{\text{a}}|^2-\big|\vec{\text{b}}\big|^2=8$
$\Rightarrow\big(8\big|\vec{\text{b}}\big|\big)^2-\big|\vec{\text{b}}\big|^2=8$ $\big[|\vec{\text{a}}|=8\big|\vec{\text{b}}\big|\big]$
$\Rightarrow64\big|\vec{\text{b}}\big|^2-\big|\vec{\text{b}}\big|^2=8$
$\Rightarrow63\big|\vec{\text{b}}\big|^2=8$
$\Rightarrow\big|\vec{\text{b}}\big|^2=\frac{8}{63}$
$\Rightarrow\big|\vec{\text{b}}\big|=\sqrt{\frac{8}{63}}$ [Magnitude of a vectoer is non-negative]
$\Rightarrow\big|\vec{\text{b}}\big|=\frac{2\sqrt{2}}{3\sqrt{7}}$
$|\vec{\text{a}}|=8\big|\vec{\text{b}}\big|=\frac{8\times2\sqrt{2}}{3\sqrt{7}}=\frac{16\sqrt{2}}{3\sqrt{7}}$
View full question & answer→Question 103 Marks
If $\vec{\text{c}}$ is perpendicular to both $\vec{\text{a}}$ and $\vec{\text{b}},$ then prove that it is perpendicular to both $\vec{\text{a}}+\vec{\text{b}}$ and $\vec{\text{a}}-\vec{\text{b}}.$
AnswerGiven that $\vec{\text{c}}$ is perpendicular to both $\vec{\text{a}}$ and $\vec{\text{b}}.$
$\Rightarrow\vec{\text{c}}.\vec{\text{a}}=0$ and $\vec{\text{c}}.\vec{\text{b}}=0\dots(1)$
Now,
$\vec{\text{c}}.\big(\vec{\text{a}}+\vec{\text{b}}\big)=\vec{\text{c}}.\vec{\text{a}}+\vec{\text{c}}.\vec{\text{b}}=0+0=0$ [From (1)]
So, $\vec{\text{c}}$ is perpendicular to $\vec{\text{a}}+\vec{\text{b}}.$
Again,
$\vec{\text{c}}.\big(\vec{\text{a}}-\vec{\text{b}}\big)=\vec{\text{c}}.\vec{\text{a}}-\vec{\text{c}}.\vec{\text{b}}=0-0=0$ [From (1)]
So, $\vec{\text{c}}$ is perpendicular to $\vec{\text{a}}-\vec{\text{b}}.$
View full question & answer→Question 113 Marks
If the sum of two unit vectors is a unit vector prove that the magnitude of their difference is $\sqrt{3.}$
AnswerGiven that $\hat{\text{a}},\hat{\text{b}}$ and $\big|\hat{\text{a}}+\hat{\text{b}}\big|$ are unit vectors.
So, $|\hat{\text{a}}|=1,|\hat{\text{b}}|=1$and $\big|\big|\hat{\text{a}}+\hat{\text{b}}\big|\big|=1$
WE have
$\big|\hat{\text{a}}+\hat{\text{b}}\big|^2+\big|\hat{\text{a}}-\hat{\text{b}}\big|^2=2\big(|\hat{\text{a}}|^2+\big|\hat{\text{b}}\big|^2\big)$
$\Rightarrow1+\big|\hat{\text{a}}-\hat{\text{b}}\big|^2=2(1+1)$
$\Rightarrow1+\big|\hat{\text{a}}-\hat{\text{b}}\big|^2=4$
$\Rightarrow\big|\hat{\text{a}}-\hat{\text{b}}\big|^2=3$
$\Rightarrow\big|\hat{\text{a}}-\hat{\text{b}}\big|=\sqrt{3}$
View full question & answer→Question 123 Marks
Show that the vectores $\vec{\text{a}}=3\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}},\vec{\text{b}}=\hat{\text{i}}-3\hat{\text{j}}+5\hat{\text{k}},\vec{\text{c}}=2\hat{\text{i}}+\hat{\text{j}}-4\hat{\text{k}}$from a right-angled triangle.
AnswerLet ABC be the given triangle and
$\overrightarrow{\text{AC}}=\vec{\text{b}}=\hat{\text{i}}-3\hat{\text{j}}+5\hat{\text{k}}$
$\overrightarrow{\text{CB}}=\vec{\text{a}}=3\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}$
$\overrightarrow{\text{AB}}=\vec{\text{c}}=2\hat{\text{i}}+\hat{\text{j}}-4\hat{\text{k}}$
$\vec{\text{a}}.\vec{\text{b}}=3+6+5=14$
$\vec{\text{b}}.\vec{\text{c}}=2-3-20=-21$
$\vec{\text{c}}.\vec{\text{a}}=6-2-4=0$
So, $\overrightarrow{\text{AB}}$ is perpendicular to $\overrightarrow{\text{CB}}.$
Thus, $\triangle\text{ABC}$ is aright-angled triangle.
View full question & answer→Question 133 Marks
If the vertices A, B and C of $\triangle\text{ABC}$ have position vectors (1, 2, 3), (-1, 0, 0) and (0, 1, 2), respectively, what is the magnitude of $\angle\text{ABC}?$
AnswerGive that
$\overrightarrow{\text{OA}}=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}};\overrightarrow{\text{OB}}=-1\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}};\overrightarrow{\text{OC}}=0\hat{\text{i}}+1\hat{\text{j}}+2\hat{\text{k}}$
$\overrightarrow{\text{AB}}=\overrightarrow{\text{OB}}-\overrightarrow{\text{OA}}=-2\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}\\\Rightarrow\Big|\overrightarrow{\text{AB}}\Big|=\sqrt{4+4+9}=\sqrt{17}$
$\overrightarrow{\text{BC}}=\overrightarrow{\text{OC}}-\overrightarrow{\text{OB}}=\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}\\\Rightarrow\Big|\overrightarrow{\text{BC}}\Big|=\sqrt{1+1+4}=\sqrt{6}$
$\overrightarrow{\text{CA}}=\overrightarrow{\text{OA}}-\overrightarrow{\text{OC}}=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\\\Rightarrow\Big|\overrightarrow{\text{CA}}\Big|=\sqrt{1+1+1}=\sqrt{3}$
$\cos\angle\text{ABC}=\frac{\big|\overrightarrow{\text{AB}}.\overrightarrow{\text{BC}}\big|}{\big|\overrightarrow{\text{AB}}\big|\big|\overrightarrow{\text{BC}}\big|}=\frac{|-2-2-6|}{(\sqrt{17})(\sqrt{6})}=\frac{10}{\sqrt{102}}$
$\Rightarrow\angle\text{ABC}=\cos^{-1}\Big(\frac{10}{\sqrt{102}}\Big)$
View full question & answer→Question 143 Marks
If $|\vec{\text{a}}|=\text{a}$ and $\big|\vec{\text{b}}\big|=\text{b},$ prove that $\Big(\frac{\vec{\text{a}}}{\text{a}^2}-\frac{\vec{\text{b}}}{\text{b}^2}\Big)^2=\Big(\frac{\vec{\text{a}}-\vec{\text{b}}}{\text{ab}}\Big)^2.$
Answer$\Big(\frac{\vec{\text{a}}}{\text{a}^2}-\frac{\vec{\text{b}}}{\text{b}^2}\Big)^2$
$=\Big|\frac{\vec{\text{a}}}{\text{a}^2}\Big|^2+\Big|\frac{\vec{\text{b}}}{\text{b}^2}\Big|^2=\frac{2\vec{\text{a}}.\vec{\text{b}}}{\text{a}^2\text{b}^2}$
$=\frac{|\vec{\text{}a}|^2}{\text{a}^4}+\frac{\big|\vec{\text{b}}\big|^2}{\text{b}^4}-\frac{2\vec{\text{a}}.\vec{\text{b}}}{\text{a}^2\text{b}^2}$
$=\frac{\text{a}^2}{\text{a}^4}+\frac{\text{b}^2}{\text{b}^4}-\frac{2\vec{\text{a}}.\vec{\text{b}}}{\text{a}^2\text{b}^2}$ (From the given information)
$=\frac{1}{\text{a}^2}+\frac{1}{\text{b}^2}-\frac{2\vec{\text{a}}.\vec{\text{b}}}{\text{a}^2\text{b}^2}$
$=\frac{\text{b}^2+\text{a}^2-2\vec{\text{a}}.\vec{\text{b}}}{\text{a}^2\text{b}^2}$
$=\frac{\text{a}^2+\text{b}^2-2\vec{\text{a}}.\vec{\text{b}}}{\text{a}^2\text{b}^2}$
$=\frac{|\vec{\text{a}}|^2+|\vec{\text{a}}|^2-2\vec{\text{a}}.\vec{\text{b}}}{\text{a}^2\text{b}^2}$ (From the given information)
$=\frac{\big(\vec{\text{a}}-\vec{\text{b}}\big)^2}{\text{a}^2\text{b}^2}$
$=\Big(\frac{\vec{\text{a}}-\vec{\text{b}}}{\text{ab}}\Big)^2$
View full question & answer→Question 153 Marks
If $\vec{\text{a}}=5\hat{\text{i}}-\hat{\text{j}}-3\hat{\text{k}}$ and $\vec{\text{b}}=\hat{\text{i}}+3\hat{\text{j}}-5\hat{\text{k}},$ then show that the vectores $\vec{\text{a}}+\vec{\text{b}}$ and $\vec{\text{a}}-\vec{\text{b}}$ are orthonal.
AnswerGiven that
$\vec{\text{a}}=5\hat{\text{i}}-\hat{\text{j}}-3\hat{\text{k}};\vec{\text{b}}=\hat{\text{i}}+3\hat{\text{j}}-5\hat{\text{k}}$
$\therefore\vec{\text{a}}+\vec{\text{b}}=5\hat{\text{i}}-\hat{\text{j}}-3\hat{\text{k}}+\hat{\text{i}}+3\hat{\text{j}}-5\hat{\text{k}}=6\hat{\text{i}}+2\hat{\text{j}}-8\hat{\text{k}}$
And $\vec{\text{a}}-\vec{\text{b}}=5\hat{\text{i}}-\hat{\text{j}}-3\hat{\text{k}}-\big(\hat{\text{i}}+3\hat{\text{j}}-5\hat{\text{k}}\big)=4\hat{\text{i}}-4\hat{\text{j}}+2\hat{\text{k}}$
Now,
$\big(\vec{\text{a}}+\vec{\text{b}}\big).\big(\vec{\text{a}}-\vec{\text{b}}\big)$
$=\big(6\hat{\text{i}}+2\hat{\text{j}}-8\hat{\text{k}}\big).\big(4\hat{\text{i}}-4\hat{\text{j}}+2\hat{\text{k}}\big)$
$=24-8-16$
$=0$
So, $\vec{\text{a}}+\vec{\text{b}}$ is orthogonal to $\vec{\text{a}}-\vec{\text{b}}.$
View full question & answer→Question 163 Marks
If $\vec{\text{a}}$ and $\vec{\text{b}}$ are two vectors of the same magnitude inclined at an angle of 30°, such that $\vec{\text{a}}.\vec{\text{b}}=3,$ find $|\vec{\text{a}}|,\big|\vec{\text{b}}\big|.$
AnswerGiven that the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$ is $30^{\circ}$
Also,
$|\vec{\text{a}}|=\big|\vec{\text{b}}\big|.$ and $\vec{\text{a}}.\vec{\text{b}}=3$
We know that
$\vec{\text{a}}.\vec{\text{b}}=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta$
$\Rightarrow3=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos30$
$\Rightarrow3=|\vec{\text{a}}|^2\Big(\frac{\sqrt{3}}{2}\Big)$
$\Rightarrow|\vec{\text{a}}|^2=\frac{6}{\sqrt{3}}=2\sqrt{3}$
$\Rightarrow|\vec{\text{a}}|=\sqrt{2\sqrt{3}}=\big|\vec{\text{b}}\big|$
$\therefore|\vec{\text{a}}|=\big|\vec{\text{b}}\big|=\sqrt{2\sqrt{3}}$
View full question & answer→Question 173 Marks
If $\vec{\text{a}}$ are $\vec{\text{b}}$ are unit vectors, then find the between $\vec{\text{a}}$ and $\vec{\text{b}},$ given that $\big(\sqrt{3}\vec{\text{a}}-\vec{\text{b}}\big)$ is aunit vector.
AnswerLet the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$ be $\theta.$
It is given that $|\vec{\text{a}}|=\big|\vec{\text{b}}\big|=\big|\vec{\text{a}}+\vec{\text{b}}\big|=1.$
$\big|\sqrt{3}\vec{\text{a}}-\vec{\text{b}}\big|=1$
$\Rightarrow\big|\sqrt{3}\vec{\text{a}}-\vec{\text{b}}\big|^2=1$
$\Rightarrow\big|\sqrt{3}\vec{\text{a}}\big|^2-2\sqrt{3}\vec{\text{a}}.\vec{\text{b}}+\big|\vec{\text{b}}\big|^2=1$
$\Rightarrow3|\vec{\text{a}}|^2-2\sqrt{3}|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta+\big|\vec{\text{b}}\big|^2=1$
$\Rightarrow3\times1-2\sqrt{3}\times1\times1\times\cos\theta+1=1$
$\Rightarrow2\sqrt{3}\cos\theta=3$
$\Rightarrow\cos\theta=\frac{\sqrt{3}}{2}=\cos\frac{\pi}{6}$
$\Rightarrow\theta=\frac{\pi}{6}$
Thus, the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$ is $\frac{\pi}{6}.$
View full question & answer→Question 183 Marks
Dot product of a vector with $\hat{\text{i}}+\hat{\text{j}}-3\hat{\text{k}},\hat{\text{i}}+3\hat{\text{j}}-2\hat{\text{k}}$ and $2\hat{\text{i}}+\hat{\text{j}}+4\hat{\text{k}}$ are 0, 5 and 8 respectively. Find the vector.
AnswerLet $\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}}$ be the required vector.
Given that
$\big(\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}}\big).\big(\hat{\text{i}}+\hat{\text{j}}-3\hat{\text{k}}\big)=0$
$\Rightarrow\text{a+b}-3\text{c}=0\dots(1)$
$\big(\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}}\big).\big(\hat{\text{i}}+3\hat{\text{j}}-2\hat{\text{k}}\big)=5$
$\Rightarrow\text{a}+3\text{b}-2\text{c}=5\dots(2)$
$\big(\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}}\big).\big(2\hat{\text{i}}+\hat{\text{j}}+4\hat{\text{k}}\big)=5$
$\Rightarrow2\text{a+b}+4\text{c}=8\dots(3)$
Solving (1), (2) and (3), we get
$\text{a}=1,\text{b}=2,\text{c}=1$
So, $\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}}=\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}$
View full question & answer→Question 193 Marks
Find the projection of $\vec{\text{b}}+\vec{\text{c}}$ on $\vec{\text{a}},$ where $\vec{\text{a}}=2\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}},\vec{\text{b}}=\hat{\text{i}}+2\hat{\text{j}}-2\hat{\text{k}}$ and $\vec{\text{c}}=2\hat{\text{i}}-\hat{\text{j}}+4\hat{\text{k}}.$
Answerprojection of $\big(\vec{\text{b}}+\vec{\text{c}}\big)$ on $\vec{\text{a}}$
$=\frac{\big(\vec{\text{b}}+\vec{\text{c}}\big).\vec{\text{a}}}{|\vec{\text{a}}|}$
$=\frac{\vec{\text{b}}.\vec{\text{a}}+\vec{\text{c}}.\vec{\text{a}}}{\sqrt{(2)^2+(-2)^2+(1)^2}}$
$=\frac{\big(\hat{\text{i}}+2\hat{\text{j}}-2\hat{\text{k}}\big)\big(2\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}\big)\big(2\hat{\text{i}}-\hat{\text{j}}+4\hat{\text{k}}\big)\big(2\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}\big)}{\sqrt{4+4+1}}$
$=\frac{(1)(2)+(2)(-2)+(-2)(1)+(2)(2)+(-1)(-2)+(4)(1)}{\sqrt{9}}$
$=\frac{2-4-2+4+2+4}{3}$
$=\frac{12-6}{3}=\frac{6}{3}=2$
projection of $\big(\vec{\text{b}}+\vec{\text{c}}\big)=2$
View full question & answer→Question 203 Marks
If $\vec{\text{a}}$ are $\vec{\text{b}}$ are two unit vectors such that $\vec{\text{a}}+\vec{\text{b}}$ is $\frac{\pi}{6}.$
AnswerLet the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$ be $\theta.$
It is given that $|\vec{\text{a}}|=\big|\vec{\text{b}}\big|=\big|\vec{\text{a}}+\vec{\text{b}}\big|=1.$
$\big|\vec{\text{a}}+\vec{\text{b}}\big|=1$
$\Rightarrow\big|\vec{\text{a}}+\vec{\text{b}}\big|^2=1$
$\Rightarrow|\vec{\text{a}}|^2+2|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta+\big|\vec{\text{b}}\big|^2=1$
$\Rightarrow1+2\times1\times1\times\cos\theta+1=1$
$\Rightarrow2\cos\theta=-1$
$\Rightarrow\cos\theta=-\frac{1}{2}=\cos\frac{2\pi}{3}$
$\Rightarrow\theta=\frac{2\pi}{3}$
View full question & answer→Question 213 Marks
If A, B and C have poition vectors (0, 1, 1), (3, 1, 5) and (0, 3, 3), respectively, show that $\triangle\text{ABC}$ is right-angled at C.
AnswerGive that
$\overrightarrow{\text{OA}}=0\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}};\overrightarrow{\text{OB}}=3\hat{\text{i}}+\hat{\text{j}}+5\hat{\text{k}};\overrightarrow{\text{OC}}=0\hat{\text{i}}+3\hat{\text{j}}+3\hat{\text{k}}$
$\overrightarrow{\text{BC}}=\overrightarrow{\text{OC}}-\overrightarrow{\text{OB}}=-3\hat{\text{i}}+2\hat{\text{j}}-2\hat{\text{k}}$
$\overrightarrow{\text{CA}}=\overrightarrow{\text{OA}}-\overrightarrow{\text{OC}}=0\hat{\text{i}}-2\hat{\text{j}}-2\hat{\text{k}}$
Now,
$\overrightarrow{\text{BC}}.\overrightarrow{\text{CA}}=0-4+4=0$
So, $\overrightarrow{\text{BC}}$ is perpendicular to $\overrightarrow{\text{CA}}$.
So, $\triangle\text{ABC}$ is right-angled at C.
View full question & answer→Question 223 Marks
Find $|\vec{\text{a}|}$ and $\big|\vec{\text{b}}\big|$ if
$\big(\vec{\text{a}}+\vec{\text{b}}\big).\big(\vec{\text{a}}-\vec{\text{b}}\big)=12$ and $|\vec{\text{a}}|=2\big|\vec{\text{b}}\big|$
AnswerHere, $\big(\vec{\text{a}}+\vec{\text{b}}\big).\big(\vec{\text{a}}-\vec{\text{b}}\big)=12$
$|\vec{\text{a}}|^2-\big|\vec{\text{b}}\big|^2=12$
$\big(2\big|\vec{\text{b}}\big|\big)^2-\big|\vec{\text{b}}\big|^2=12$ $\big[\text{Using|}\vec{\text{a}}|=2\big|\vec{\text{b}}\big|\big]$
$4\big|\vec{\text{b}}\big|^2-\big|\vec{\text{b}}\big|^2=12$
$3\big|\vec{\text{b}}\big|^2=12$
$\big|\vec{\text{b}}\big|^2=\frac{12}{3}$
$\big|\vec{\text{b}}\big|^2=4$
$\big|\vec{\text{b}}\big|=2$
$\big|\vec{\text{a}}\big|=2\big|\vec{\text{b}}\big|=2(2)$
$\big|\vec{\text{a}}\big|=4$
$\big|\vec{\text{b}}\big|=2$
View full question & answer→Question 233 Marks
If $\vec{\text{a}}=2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}},\vec{\text{b}}=\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}}$ and $\vec{\text{c}}=\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}}.$ find $\lambda$ such that $\vec{\text{a}}$ is perpendicular to $\lambda\vec{\text{b}}+\vec{\text{c}}.$
AnswerThe given vectors are $\vec{\text{a}}=2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}},\vec{\text{b}}=\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}}$ and $\vec{\text{c}}=\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}}.$
Now,
$\lambda\vec{\text{b}}+\vec{\text{c}}=\lambda(\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}})+(\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}})\\=(\lambda+1)\hat{\text{i}}+(\lambda+3)\hat{\text{j}}-(2\lambda+1)\hat{\text{k}}$
It is given
$\vec{\text{a}}\perp\big(\lambda\vec{\text{b}}+\vec{\text{c}}\big)$
$\Rightarrow\vec{\text{a}}.\big(\lambda\vec{\text{b}}+\vec{\text{c}}\big)=0$
$\Rightarrow\big(2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}\big).\Big[(\lambda+1)\hat{\text{i}}+(\lambda+3)\hat{\text{j}}-(2\lambda+1)\hat{\text{k}}\Big]=0$
$\Rightarrow2(\lambda+1)-(\lambda+3)-(2\lambda+1)=0$
$\Rightarrow2\lambda+2-\lambda-3-2\lambda-1=0$
Thus, the value of $\lambda$ is -2
View full question & answer→Question 243 Marks
Write the projection of $\vec{\text{b}}+\vec{\text{c}}$ on $\vec{\text{a}}$ when $\vec{\text{a}}=2\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}},\vec{\text{b}}=\hat{\text{i}}+2\hat{\text{j}}-2\hat{\text{k}}$ and $\vec{\text{c}}=2\hat{\text{i}}-\hat{\text{j}}+4\hat{\text{k}}.$
AnswerGiven that
$\vec{\text{a}}=2\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}$
$\vec{\text{b}}=\hat{\text{i}}+2\hat{\text{j}}-2\hat{\text{k}}$
$\vec{\text{c}}=2\hat{\text{i}}-\hat{\text{j}}+4\hat{\text{k}}$
$\vec{\text{b}}+\vec{\text{c}}=\hat{\text{i}}+2\hat{\text{j}}-2\hat{\text{k}}+2\hat{\text{i}}-\hat{\text{j}}+4\hat{\text{k}}$
$=3\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}$
Projection of $\vec{\text{b}}+\vec{\text{c}}$ on $\vec{\text{a}}$ is
$\frac{\big(\vec{\text{b}}+\vec{\text{c}}\big).\vec{\text{a}}}{|\vec{\text{a}}|}$
$=\frac{\big(3\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}\big).\big(2\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}\big)}{2\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}}$
$=\frac{6-2+2}{\sqrt{4+4+1}}$
$=\frac{6}{3}$
$=2$
View full question & answer→Question 253 Marks
If $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are three non-coplanar vectors, such that $\vec{\text{d}}.\vec{\text{a}}=\vec{\text{d}}.\vec{\text{b}}=\vec{\text{d}}.\vec{\text{c}}=0,$ then show that $\vec{\text{d}}$ is the null vector.
AnswerGiven that
$\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are three non-coplanar vectors such that
$\vec{\text{d}}.\vec{\text{a}}=\vec{\text{d}}.\vec{\text{b}}=\vec{\text{d}}.\vec{\text{c}}=0$
Given that
$\vec{\text{d}}.\vec{\text{a}}=0$
⇒ $\vec{\text{d}}$ perpendicular to $\vec{\text{a}}$
or $\vec{\text{d}}=0\dots(1)$
$\vec{\text{d}}.\vec{\text{b}}=0$
⇒ $\vec{\text{d}}$ is perpendicular to $\vec{\text{b}}$ or $\vec{\text{d}}=0\dots(2)$
$\vec{\text{d}}.\vec{\text{c}}=0$
⇒ $\vec{\text{d}}$ is perpendicular to $\vec{\text{c}}$ or $\vec{\text{d}}=0\dots(3)$
From (1), (2), (3), we get
$\vec{\text{d}}$ is perpendicular to $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ or $\vec{\text{d}}=0,$ but $\vec{\text{d}}$ can not be perpendicular to $\vec{\text{a}},\vec{\text{b}}$ and $\vec{\text{c}},$ because $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are three non-coplanar vectors, so
$\vec{\text{d}}=0,$
View full question & answer→Question 263 Marks
Find $|\vec{\text{a}|}$ and $\big|\vec{\text{b}}\big|$ if
$\big(\vec{\text{a}}+\vec{\text{b}}\big).\big(\vec{\text{a}}-\vec{\text{b}}\big)=3$ and $|\vec{\text{a}}|=2\big|\vec{\text{b}}\big|$
AnswerHere, $\big(\vec{\text{a}}+\vec{\text{b}}\big).\big(\vec{\text{a}}-\vec{\text{b}}\big)=3$ $|\vec{\text{a}}|^2-\big|\vec{\text{b}}\big|^2=3$ $\big(2\big|\vec{\text{b}}\big|\big)^2-\big|\vec{\text{b}}\big|^2=3$ $\big[\text{using} |\vec{\text{a}}|=2\big|\vec{\text{b}}\big|\big]$ $4\big|\vec{\text{b}}\big|^2-\big|\vec{\text{b}}\big|^2=3$$3\big|\vec{\text{b}}\big|^2=3$
$\big|\vec{\text{b}}\big|^2=\frac{3}{3}$
$\big|\vec{\text{b}}\big|^2=1$
$\big|\vec{\text{b}}\big|=1$
$|\vec{\text{a}}|=2\big|\vec{\text{b}}\big|$ $=2(1)$ $|\vec{\text{a}}|=2$ $\big|\vec{\text{b}}\big|=1$
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