Question 513 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 523 Marks
Find a vector of magnitude 5 units, and parallel to the resultant of the vectors $\vec{\text{a}}=2\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}}$ and $\vec{\text{b}}=\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}.$
AnswerWe have,
$\vec{\text{a}}=2\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}}\ \text{and}\ \vec{\text{b}}=\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}.$
$\text{Let}\ \vec{\text{c}}\ \text{be the resultant of}\ \vec{\text{a}}\ \text{and}\ \vec{\text{b}}$
Then,
$\vec{\text{c}}=\vec{\text{a}}+\vec{\text{b}}=(2+1)\hat{\text{i}}+(3-2)\hat{\text{j}}+(-1+1)\hat{\text{k}}=3\hat{\text{i}}+\hat{\text{ j}}$
$\therefore|\vec{\text{c}}|=\sqrt{3^{2}+1^{2}}=\sqrt{9+1}=\sqrt{10}$
$\therefore\hat{\text{c}}=\frac{\vec{\text{c}}}{|\vec{\text{c}}|}=\frac{\big(3\hat{\text{i}}+\hat{\text{j}}\big)}{\sqrt{10}}$
Hence, the vector of magnltude 5 units and parallel to the resultant of vectors $\vec{\text{a}}\ \text{and}\ \vec{\text{b}} \ \text{is}$
$\pm5\cdot\hat{c}=\pm5\cdot\frac{1}{\sqrt{10}}\big(3\hat{\text{i}}+\hat{\text{j}}\big)$ $=\pm\frac{3\sqrt{10}\hat{\text{i}}}{2}\pm\frac{\sqrt{10}}{2}\hat{\text{j}}.$
View full question & answer→Question 533 Marks
If D, E, F are the mid-points of side BC, CA and AB respectively of a triangle ABC, write the value of $\overrightarrow{\text{AD}}+\overrightarrow{\text{BE}}+\overrightarrow{\text{CF}}$.
AnswerGiven: D, E, F are the mid-points of the sides BC, CA, AB respectively. Then, the position vectors of the mid-points D, E, F are given by $\frac{\vec{\text{b}}+\vec{\text{c}}}2,\ \frac{\vec{\text{c}}+\vec{\text{a}}}2,\ \frac{\vec{\text{a}}+\vec{\text{b}}}2$Now, $\overrightarrow{\text{AD}}+\overrightarrow{\text{BE}}+\overrightarrow{\text{CF}}$
$=\Big(\frac{\vec{\text{b}}+\vec{\text{c}}}2\Big)-\vec{\text{a}}+\Big(\frac{\vec{\text{c}}+\vec{\text{a}}}2\Big)-\vec{\text{b}}+\Big(\frac{\vec{\text{a}}+\vec{\text{b}}}2\Big)-\vec{\text{c}}$
$=2\Big(\frac{\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}}2\Big)-\big(\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big)$
$=\big(\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big)-\big(\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big)$
$=\vec0$
View full question & answer→Question 543 Marks
Find a vector $\vec{\text{r}}$ of magnitude $3\sqrt{2}$ units which makes an angle of $\frac{\pi}{4}$ and $\frac{\pi}{2}$ with and z-axes respectively.
AnswerSuppose vector $\vec{\text{r}}$ makes an angle with the $\alpha$ x-axis.Let l, m, n be the direction cosines of $\vec{\text{r}}$. Then,
$\text{l}=\cos\alpha,\text{m}=\cos\frac{\pi}{4}=\frac{1}{\sqrt{2}},\text{n}=\cos\frac{\pi}{2}=0$
Now,
$\text{l}^2+\text{m}^2+\text{n}^2=1$
$\Rightarrow\cos^2\alpha+\frac{1}{2}+0=1$
$\Rightarrow\cos^2\alpha=1-\frac{1}{2}=\frac{1}{2}$
$\Rightarrow\cos^2\alpha=\pm\frac{1}{\sqrt{2}}$
We know that,
$\vec{\text{r}}=|\vec{\text{r}}|\big(\text{l}\hat{\text{i}}+\text{m}\hat{\text{j}}+\text{n}\hat{\text{k}}\big)$
$\therefore\vec{\text{r}}=3\sqrt{2}\Big(\pm\frac{1}{\sqrt{2}}\hat{\text{i}}+\frac{1}{\sqrt{2}}\hat{\text{i}}+0\hat{\text{k}}\Big)\ (|\vec{\text{a}}|=3\sqrt{2})$
$\Rightarrow\vec{\text{r}}=\pm3\hat{\text{i}}+3\hat{\text{j}}$
View full question & answer→Question 553 Marks
Show that the points A, B and C with position vectors, $\vec{a}=3\hat{i}-4\hat{j}-4\hat{k},\ \vec{b}=2\hat{i}-\hat{j}+\hat{k}\ $ $\text{and}\ \vec{c}=\hat{i}-3\hat{j}-5\hat{k},$ respectively form the vertices of a right angled triangle.
AnswerPosition vectors of points A, B, and C are respectively given as:
$\vec{a}=3\hat{i}-4\hat{j}-4\hat{k},\ \vec{b}=2\hat{i}-\hat{j}+\hat{k}\ \text{and}\ \vec{c}$ $=\hat{i}-3\hat{j}-5\hat{k}$
$\therefore\overrightarrow{\text{AB}}=\vec{b}-\vec{a}={(2-3)}\hat{i}+(-1+4)\hat{j}+(1+4)\hat{k}$ $=-\hat{i}+3\hat{j}+5\hat{k}$
$\overrightarrow{\text{BC}}=\vec{c}-\vec{b}={(1-2)}\hat{i}+(-3+1)\hat{j}+(-5-1)\hat{k}$ $=-\hat{i}-2\hat{j}-6\hat{k}$
$\overrightarrow{\text{CA}}=\vec{a}-\vec{c}={(3-1)}\hat{i}+(-4+3)\hat{j}+(-4+5)\hat{k}$ $=2\hat{i}-\hat{j}+\hat{k}$
$\therefore\bigg|\overrightarrow{\text{AB}}\bigg|^2=(-1)^2+3^2+5^2=1+9+25=35$
$\bigg|\overrightarrow{\text{BC}}\bigg|^2=(-1)^2+(-2)^2+(-6)^2$ $=1+4+36=41$
$\bigg|\overrightarrow{\text{CA}}\bigg|^2=2^2+(-1)^2+1^2=4+1+1=6$
$\therefore\bigg|\overrightarrow{\text{AB}}\bigg|^2+\bigg|\overrightarrow{\text{CA}}\bigg|^2=36+6=41=\bigg|\overrightarrow{\text{BC}}\bigg|^2$
Hence, ABC is a right-angled triangle.
View full question & answer→Question 563 Marks
Find the area of the parallelogram whose diagonals are:
$2\hat{\text{i}}+\hat{\text{k}}$ and $\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$
AnswerGiven, $\text{d}_1=2\hat{\text{i}}+\hat{\text{k}}$
$\text{d}_2=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$
$\vec{\text{d}_1}\times\vec{\text{d}_2}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\2&0&1\\1&1&1 \end{vmatrix}$
$=\hat{\text{i}}(0-1)-\hat{\text{j}}(2-1)+\hat{\text{k}}(2-0)$
$=-\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}$
$\big|\vec{\text{d}}_1\times\vec{\text{d}}_2\big|=\sqrt{(-1)^2+(-1)^2+(2)^2}$
$=\sqrt{1+1+4}$
$=\sqrt{6}$
Area of parallelogram $=\frac{1}{2}\big|\vec{\text{d}}_1\times\vec{\text{d}}_2\big|$
Area of parallelogram $=\frac{1}{2}\sqrt{6}\text{ sq. unit}$
View full question & answer→Question 573 Marks
Find the unit vector in the direction of vector $\overrightarrow{\text{PQ}}$, where P and Q are the points (1, 2, 3) and (4, 5, 6).
AnswerLet $\vec{\text{a}}\text{ and }\vec{\text{b}}$ are the position vectors of the point P(1, 2, 3) and Q(4, 5, 6)
Then,
$\vec{\text{a}}=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$
$\vec{\text{b}}=4\hat{\text{i}}+5\hat{\text{j}}+6\hat{\text{k}}$
So,
$\overrightarrow{\text{PQ}}=\vec{\text{b}}-\vec{\text{a}}$
$=4\hat{\text{i}}+5\hat{\text{j}}+6\hat{\text{k}}-\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}$
$=3\hat{\text{i}}+3\hat{\text{j}}+3\hat{\text{k}}$
Now, $\Big|\overrightarrow{\text{PQ}}\Big|=\sqrt{3^2+3^2+3^2}$
$=\sqrt{9+9+9}$
$=3\sqrt3$
Therefore, Unit vector parallel to $\overrightarrow{\text{PQ}}=\frac{\overrightarrow{\text{PQ}}}{|\text{PQ}|}=\frac{1}{3\sqrt3}\big(3\hat{\text{i}}+3\hat{\text{j}}+3\hat{\text{k}}\big)$
$=\frac{1}{\sqrt3}\big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)$
View full question & answer→Question 583 Marks
A vector $\vec{\text{r}}$ is inclined to x-axis at 45º and y-axis at 60º. If $|\vec{\text{r}}|=8$ units, find $\vec{\text{r}}$.
AnswerHere, $\alpha=45^{\circ},\ \beta=60^{\circ},\ \gamma=\theta$ (say)
$\text{l}=\cos\alpha$
$=\cos45^{\circ}$
$\text{l}=\frac{1}{\sqrt2}$
$\text{m}=\cos\beta$
$=\cos60^{\circ}$
$\text{m}=\frac{1}2$
$\text{n}=\cos\theta$
Put l, m and n in,
$\text{l}^2+\text{m}^2+\text{n}^2=1$
$\Big(\frac{1}{\sqrt2}\Big)^2+\Big(\frac{1}2\Big)^2+\cos^2\theta=1$
$\frac{2+1}4+\cos^2\theta=1$
$\frac{3}4+\cos^2\theta=1$
$\cos^2\theta=\frac{1}1-\frac{3}4$
$=\frac{4-3}4$
View full question & answer→Question 593 Marks
If either $\vec{\text{a}}=\vec{0}$ or $\vec{\text{b}}=\vec{0},$ then $\vec{\text{a}}\times\vec{\text{b}}=\vec{0}.$ is the converse true? justify your answer with an example.
AnswerIf $\vec{\text{a}}=\vec{0}$ or $\vec{\text{b}}=\vec{0},$ then $|\vec{\text{a}}|\big|\vec{\text{b}}\big|\sin\theta\text{ n}=\vec{0}.$
$\Rightarrow\vec{\text{a}}\times\vec{\text{b}}=\vec{0}$
But the converse is not true as whenever $\vec{\text{a}}\times\vec{\text{b}}=\vec{0},$ we cannot be sure that either $\vec{\text{a}}=\vec{0}$ or $\vec{\text{b}}=\vec{0}.$
For exampale:
$\vec{\text{a}}=\hat{\text{i}}+2\hat{\text{j}}+3\vec{\text{k}}$
$\vec{\text{b}}=\hat{\text{i}}+2\hat{\text{j}}+3\vec{\text{k}}$
Here,
$\vec{\text{a}}\neq0$
$\vec{\text{b}}\neq0$
But $\vec{\text{a}}\times\vec{\text{b}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\1&2&3\\1&2&3 \end{vmatrix}$
$=0\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}}$
$=\vec{0}$
View full question & answer→Question 603 Marks
Find the area of the parallelogram determinrd by the vectors:
$\hat{\text{i}}-3\hat{\text{j}}+\hat{\text{k}}$ and $\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}.$
AnswerLet:
$\vec{\text{a}}=\hat{\text{i}}-3\hat{\text{j}}+\hat{\text{k}}$
$\vec{\text{b}}=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$
$\vec{\text{a}}\times\vec{\text{b}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\1&-3&1\\1&1&1 \end{vmatrix}$
$=(-3-1)\hat{\text{i}}-(1-1)\hat{\text{j}}+(1+3)\hat{\text{k}}$
$=-4\hat{\text{i}}+0\hat{\text{j}}+4\hat{\text{k}}$
Area of the parallelogram $=\big|\vec{\text{a}}\times\vec{\text{b}}\big|$
$=\sqrt{(-4)^2+0+4^2}$
$=\sqrt{32}$
$=4\sqrt{2}\text{ sq. units.}$
View full question & answer→Question 613 Marks
If $\vec{\text{a}}=3\hat{\text{i}}-\hat{\text{j}}-2\hat{\text{k}}$ and $\vec{\text{b}}=2\hat{\text{i}}+3\hat{\text{j}}+\hat{\text{k}},$ find $\big(\vec{\text{a}}+2\vec{\text{b}}\big)\times\big(\vec{\text{a}}-2\vec{\text{b}}\big).$
AnswerGiven: $\vec{\text{a}}=3\hat{\text{i}}-\hat{\text{j}}-2\hat{\text{k}}$ $\vec{\text{b}}=2\hat{\text{i}}+3\hat{\text{j}}+\hat{\text{k}}$$\therefore\vec{\text{a}}+2\vec{\text{b}}=3\hat{\text{i}}-\hat{\text{j}}-2\hat{\text{k}}+2\big(2\hat{\text{i}}+3\hat{\text{j}}+\hat{\text{k}}\big)$
$=7\hat{\text{i}}+5\hat{\text{j}}+0\hat{\text{k}}$
$\therefore2\vec{\text{a}}-\vec{\text{b}}\big(3\hat{\text{i}}-\hat{\text{j}}-2\hat{\text{k}}\big)-\big(2\hat{\text{i}}+3\hat{\text{j}}+\hat{\text{k}}\big)$
$=4\hat{\text{i}}-5\hat{\text{j}}-5\hat{\text{k}}$
$\big(\vec{\text{a}}+2\vec{\text{b}}\big)\times\big(2\vec{\text{a}}-\vec{\text{b}}\big)=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\7&5&0\\4&-5&-5 \end{vmatrix}$
$=\hat{\text{i}}(-25+0)-\hat{\text{j}}(-35+0)+\hat{\text{k}}(-35-20)$
$=-25\hat{\text{i}}+35\hat{\text{j}}-55\hat{\text{k}}$
View full question & answer→Question 623 Marks
$\text{If}\ \ \vec{\text{a}}=\vec{\text{b}}+\vec{\text{c}},$ then is it true that $\big|\vec{\text{a}}\big|=\big|\vec{\text{b}}\big|+\big|\vec{\text{c}}\big|?$ Justify your answer.
Answer$\text{If}\ \triangle\text{ABC},\ \text{let}\ \overrightarrow{\text{CB}}=\vec{\text{a}},\ \overrightarrow{\text{CA}}=\vec{\text{b}},\text{and}\ \overrightarrow{\text{AB}}=\vec{\text{c}}$ (as shown in the following figure).
Now, by the triangle law of vector additio, we have $\vec{\text{a}}=\vec{\text{b}}+\vec{\text{c}}$It is clearly known that $|\vec{\text{a}}|,\big|\vec{\text{b}}\big|,\ \text{and}\ \big|\vec{\text{c}}\big|$ represent the sides of $\triangle\text{ABC}.$
Also, it is known that the sum of the lengths of any two sides of a triangle is greater than the third side. $\therefore\ |\vec{\text{a}}|<\big|\vec{b}\big|+|\vec{c}|$ Hence, it is not true that $|\vec{\text{a}}|=\big|\vec{\text{b}}\big|+\big|\vec{\text{c}}\big|$ View full question & answer→Question 633 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 643 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 653 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 663 Marks
Find the position vector of the mid-point of the vector joining the points $\text{P}\big(2\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}}\big)$ and $\text{Q}\big(4\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}}\big)$.
AnswerGiven: $\text{P}\big(2\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}}\big)$ and $\text{Q}\big(4\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}}\big)$The position vector of the mid-point of the vector
joining these points $=\frac{\text{Position vector of P}+\text{Position vector of Q}}{2}$
$=\frac{\big(2\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}}\big)+\big(4\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}}\big)}{2}$
$=\frac{6\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}}}{2}$
$=3\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$
View full question & answer→Question 673 Marks
If $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are three unit vectors such that $\vec{\text{a}}\times\vec{\text{b}}=\vec{\text{c}},\vec{\text{b}}\times\vec{\text{c}}=\vec{\text{a}},\vec{\text{c}}\times\vec{\text{a}}=\vec{\text{b}}.$Show that $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ from an orthonormal right handed triad of unit vectors.
AnswerGiven:
$\vec{\text{a}}\times\vec{\text{b}}=\vec{\text{c}}$
$\vec{\text{b}}\times\vec{\text{c}}=\vec{\text{a}}$
$\vec{\text{c}}\times\vec{\text{a}}=\vec{\text{b}}\dots(1)$
Now,
$\big|\vec{\text{a}}\times\vec{\text{b}}\big|=|\vec{\text{c}}|=1$ ( $\because\vec{\text{c}}$ is a unit vector)
$\big|\vec{\text{b}}\times\vec{\text{c}}\big|=|\vec{\text{a}}|=1$ ($\because\vec{\text{a}}$ is a unit vector)
$\big|\vec{\text{c}}\times\vec{\text{a}}\big|=\big|\vec{\text{b}}\big|=1$ ($\because\vec{\text{b}}$ is a unit vector)
$\therefore\big|\vec{\text{a}}\times\vec{\text{b}}\big|=\big|\vec{\text{b}}\times\vec{\text{c}}\big|=\big|\vec{\text{c}}\times\vec{\text{a}}\big|=1\dots(2)$
From (1) and (2), we know
$\vec{\text{a}},\vec{\text{b}}$ and $\vec{\text{c}}$ form an orthonormal right handed triad of unit vectors.
View full question & answer→Question 683 Marks
For any two vectors $\vec{\text{a}}$ and $\vec{\text{b}},$ prove that $\big|\vec{\text{a}}\times\vec{\text{b}}\big|^2=\begin{vmatrix}\vec{\text{a}}.\vec{\text{a}}&\vec{\text{a}}.\vec{\text{b}}\\\vec{\text{b}}.\vec{\text{a}}&\vec{\text{b}} .\vec{\text{b}}\end{vmatrix}.$
Answer$\text{RHS}=\begin{vmatrix}\vec{\text{a}}.\vec{\text{a}}&\vec{\text{a}}.\vec{\text{b}}\\\vec{\text{b}}.\vec{\text{a}}&\vec{\text{b}} .\vec{\text{b}}\end{vmatrix}.$
$=\begin{vmatrix}|\vec{\text{a}}|^2&|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta\\|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta&\big|\vec{\text{b}}\big|^2 \end{vmatrix}$
$=|\vec{\text{a}}|^2\big|\vec{\text{b}}\big|^2-|\vec{\text{a}}|^2\big|\vec{\text{b}}\big|^2\cos^2\theta$
$=|\vec{\text{a}}|^2\big|\vec{\text{b}}\big|^2(1-\cos^2\theta)$
$=|\vec{\text{a}}|^2\big|\vec{\text{b}}\big|^2\sin^2\theta$
$=\big(|\vec{\text{a}}|\big|\vec{\text{b}}\big|\sin\theta\big)^2$
$=\big|\vec{\text{a}}\times\vec{\text{b}}\big|^2$
$=\text{LHS}$
Hence proved.
View full question & answer→Question 693 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 703 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 713 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 723 Marks
Find a vactor of magnitude $\sqrt{171}$ which is perpendicular to both of the vectors $\vec{\text{a}}=\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}$ and $\vec{\text{b}}=3\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}.$
AnswerThe Given vectors are $\vec{\text{a}}=\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}$ and $\vec{\text{b}}=3\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}$
Unit vectors perpenticular to both $\vec{\text{a}}$ and $\vec{\text{b}}=\pm\frac{\vec{\text{a}}\times\vec{\text{b}}}{\big|\vec{\text{a}}\times\vec{\text{b}}\big|}$
Now,
$\vec{\text{a}}\times\vec{\text{b}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\1&2&-3\\3&-1&2\end{vmatrix}=\hat{\text{i}}-11\hat{\text{j}}-7\hat{\text{k}}$
$\therefore\big|\vec{\text{a}}\times\vec{\text{b}}\big|=\big|\hat{\text{i}}-11\hat{\text{j}}-7\hat{\text{k}}\big|$
$=\sqrt{1^2+(-11)^2+(-7)^2}$
$=\sqrt{1+121+49}$
$=\sqrt{171}$
Unit vectors perpendicular to both $\vec{\text{a}}$ and $\vec{\text{b}}=\pm\frac{\hat{\text{i}}-11\hat{\text{j}}-7\hat{\text{k}}}{\sqrt{171}}$
Required vectors $=\sqrt{171}\Big(\pm\frac{\hat{\text{i}}-11\hat{\text{j}}-7\hat{\text{k}}}{\sqrt{171}}\Big)=\pm\big(\hat{\text{i}}-11\hat{\text{j}}-7\hat{\text{k}}\big)$
Thus, the vectors of magnitude $\sqrt{171}$ Which are perpendicular to both the given vector are $\pm\big(\hat{\text{i}}-11\hat{\text{j}}-7\hat{\text{k}}\big).$
View full question & answer→Question 733 Marks
What inference can you draw if $\vec{\text{a}}\times\vec{\text{b}}=\vec{0}$ and $\vec{\text{a}}.\vec{\text{b}}=0.$
AnswerGiven:
$\big|\vec{\text{a}}\times\vec{\text{b}}\big|=\vec{0}$
$\Rightarrow\vec{\text{a}}=0$
$\vec{\text{b}}=0$
$\therefore\vec{\text{a}}||\vec{\text{b}}$
Also,
$\vec{\text{a}}.\vec{\text{b}}=0$
$\Rightarrow|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta=0$
$\Rightarrow\vec{\text{a}}=\vec{0}$ or $\vec{\text{b}}=\vec{0}$ or, $\vec{\text{a}}\perp\vec{\text{b}}$
But $\vec{\text{a}}$ cannot be both perpendicular as well as parallel to $\vec{\text{b}}.$
$\therefore|\vec{\text{a}}|=0$
$\big|\vec{\text{b}}\big|=0$
View full question & answer→Question 743 Marks
Show that the points whose position vectors are as given below are collinear:
$3\hat{\text{i}}-2\hat{\text{j}}+4\hat{\text{k}},\ \hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$ and $-\hat{\text{i}}+4\hat{\text{j}}-2\hat{\text{k}}$
AnswerLet the points be A, B and C with position vectors $3\hat{\text{i}}-2\hat{\text{j}}+4\hat{\text{k}},\ \hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$ and $-\hat{\text{i}}+4\hat{\text{j}}-2\hat{\text{k}}$ respectively. Then, $\overrightarrow{\text{AB}}=$ Position vector of B - Position vector of A$=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}-3\hat{\text{i}}+2\hat{\text{j}}-4\hat{\text{k}}$
$=-2\hat{\text{i}}+3\hat{\text{j}}-3\hat{\text{k}}$
$\overrightarrow{\text{BC}}=$ Position vector of C - Position vector of B
$=-\hat{\text{i}}+4\hat{\text{j}}-2\hat{\text{k}}-\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}}$
$=-2\hat{\text{i}}+3\hat{\text{j}}-3\hat{\text{k}}$
$\therefore\ \overrightarrow{\text{AB}}=\overrightarrow{\text{BC}}$
So, $\overrightarrow{\text{AB}}$ and $\overrightarrow{\text{BC}}$ are parallel vectors. But B is a point common to them. Hence, A, B, and C are collinear.
View full question & answer→Question 753 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 763 Marks
Find a vector of magnitude of 5 units parallel to the resultant of the vectors $\vec{\text{a}}=2\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}}$ and $\vec{\text{b}}=\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}$.
AnswerGiven that, $\vec{\text{a}}=2\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}}$ and $\vec{\text{b}}=\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}$Thus, Find a vector of mangnitude of 5 units parallel to the resultant of the vectors
$\vec{\text{a}}+\vec{\text{b}}=2\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}}+\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}$
$\Rightarrow\vec{\text{a}}+\vec{\text{b}}=3\hat{\text{i}}+\hat{\text{j}}$
$\Rightarrow\big|\vec{\text{a}}+\vec{\text{b}}\big|=\sqrt{9+1}=\sqrt{10}$
Thus, the unit vector along the resultant vector $\vec{\text{a}}+\vec{\text{b}}$ is $\frac{3\hat{\text{i}}+\hat{\text{j}}}{\sqrt{10}}$
The vector of magnitude of 5 units parallel to the resultant vector$=\frac{3\hat{\text{i}}+\hat{\text{j}}}{\sqrt{10}}\times5=\sqrt{\frac{5}{2}}\big(3\hat{\text{i}}+\hat{\text{j}}\big)$
View full question & answer→Question 773 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 783 Marks
If the vertices A, B, C of a triangle ABC are the points with position vectors $\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}},\ \text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}},\ \text{c}_1\hat{\text{i}}+\text{c}_2\hat{\text{j}}+\text{c}_3\hat{\text{k}}$ respectively, what are the vectors determined by its sides? Find the length of these vectors.
AnswerGiven the vertices of a triangle A, B and C with position vectors $\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}},\ \text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}$ and $\text{c}_1\hat{\text{i}}+\text{c}_2\hat{\text{j}}+\text{c}_3\hat{\text{k}}$ respectively. Then,
$\overrightarrow{\text{AB}}=(\text{b}_1-\text{a}_1)\hat{\text{i}}+(\text{b}_2-\text{a}_2)\hat{\text{j}}+(\text{b}_3-\text{a}_3)\hat{\text{k}}$
$\overrightarrow{\text{BC}}=(\text{c}_1-\text{b}_1)\hat{\text{i}}+(\text{c}_2-\text{b}_2)\hat{\text{j}}+(\text{c}_3-\text{b}_3)\hat{\text{k}}$
$\overrightarrow{\text{CA}}=(\text{a}_1-\text{c}_1)\hat{\text{i}}+(\text{a}_2-\text{c}_2)\hat{\text{j}}+(\text{a}_3-\text{c}_3)\hat{\text{k}}$
Therefore, the length of these vectors are:
$\Big|\overrightarrow{\text{AB}}\Big|=\sqrt{(\text{b}_1-\text{a}_1)^2+(\text{b}_2-\text{a}_2)^2+(\text{b}_3-\text{a}_3)^2}$
$\Big|\overrightarrow{\text{BC}}\Big|=\sqrt{(\text{c}_1-\text{b}_1)^2+(\text{c}_2-\text{b}_2)^2+(\text{c}_3-\text{b}_3)^2}$
$\Big|\overrightarrow{\text{CA}}\Big|=\sqrt{(\text{a}_1-\text{c}_1)^2+(\text{a}_2-\text{c}_2)^2+(\text{a}_3-\text{c}_3)^2}$
View full question & answer→Question 793 Marks
Give a condition that three vectors $\vec{\text{a}},\ \vec{\text{b}}\text{ and }\vec{\text{c}}$ from the three sides of a triangle. what are the other possibilities?
AnswerLet ABC be a triangle such that $\overrightarrow{\text{BC}}=\vec{\text{a}},\ \overrightarrow{\text{AB}}=\vec{\text{c}}\text{ and }\overrightarrow{\text{CA}}=\vec{\text{b}}$. Then,
$\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}=\overrightarrow{\text{BC}}+\overrightarrow{\text{CA}}+\overrightarrow{\text{AB}}$
$\Rightarrow\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}=\overrightarrow{\text{BA}}+\overrightarrow{\text{AB}}$ $\big[\because\ \overrightarrow{\text{BC}}+\overrightarrow{\text{CA}}=\overrightarrow{\text{BA}}\big]$
$\Rightarrow\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}=\overrightarrow{\text{BB}}$ [Using triangle law]
$\Rightarrow\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}=\vec0$ [By defination of null vector]
Other possibilities are,
- $\vec{\text{c}}+\vec{\text{a}}=\vec{\text{b}}$
- $\vec{\text{a}}+\vec{\text{b}}=\vec{\text{c}}$
- $\vec{\text{b}}+\vec{\text{c}}=\vec{\text{a}}$
View full question & answer→Question 803 Marks
Find the area of the parallelogram whose diagonals are:
$3\hat{\text{i}}+4\hat{\text{j}}$ and $\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$
AnswerGiven, $\text{d}_1=3\hat{\text{i}}+4\hat{\text{j}}$
$\text{d}_2=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$
$\vec{\text{d}_1}\times\vec{\text{d}_2}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\3&4&0\\1&1&1 \end{vmatrix}$
$=\hat{\text{i}}(4-0)-\hat{\text{j}}(3-0)+\hat{\text{k}}(3-4)$
$=4\hat{\text{i}}-3\hat{\text{j}}-\hat{\text{k}}$
$\big|\vec{\text{d}}_1\times\vec{\text{d}}_2\big|=\sqrt{(4)^2+(-3)^2+(-1)^2}$
$=\sqrt{16+9+1}$
$=\sqrt{26}$
Area of parallelogram $=\frac{1}{2}\big|\vec{\text{d}}_1\times\vec{\text{d}}_2\big|$
Area of parallelogram $=\frac{\sqrt{26}}{2}\text{ sq. unit}$
View full question & answer→Question 813 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 823 Marks
Find a vector of magnitude 49, which is perpendicular to both the vectors $2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}}$ and $3\hat{\text{i}}-6\hat{\text{j}}+2\hat{\text{k}}.$
AnswerLet, $\vec{\text{a}}=2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}},\vec{\text{b}}=3\hat{\text{i}}-6\hat{\text{j}}+2\hat{\text{k}}$
If $\vec{\text{a}}=\text{a}_1\hat{\text{i}}+\text{b}_1\hat{\text{j}}+\text{c}_1\hat{\text{k}}$ and
$\vec{\text{b}}=\text{a}_2\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{c}_2\hat{\text{k}},$ then,
$\vec{\text{a}}\times\vec{\text{b}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\\text{a}_1&\text{b}_1&\text{c}_1\\\text{a}_2&\text{b}_2&\text{c}_2 \end{vmatrix}$
$=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\2&3&6\\3&-6&2 \end{vmatrix}$
$=\hat{\text{i}}(6+36)-\hat{\text{j}}(4-18)+\hat{\text{k}}(-12-9)$
$=42\hat{\text{i}}+14\hat{\text{j}}-21\hat{\text{k}}$
$=7\big(6\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}\big)$
$\big|\vec{\text{a}}\times\vec{\text{b}}\big|=7\sqrt{(6)^2+(2)^2(-3)^2}$
$=7\sqrt{36+4+9}$
$\big|\vec{\text{a}}\times\vec{\text{b}}\big|=7\sqrt{49}$
$\big|\vec{\text{a}}\times\vec{\text{b}}\big|=7\times7$
$\big|\vec{\text{a}}\times\vec{\text{b}}\big|=49$
View full question & answer→Question 833 Marks
If D is the mid-point of side BC of a triangle ABC such that $\overrightarrow{\text{AB}}+\overrightarrow{\text{AC}}=\lambda\overrightarrow{\text{AD}}$, write the value of $\lambda$.
AnswerGiven: D is the mid-point of the side BC of a triangle ABC such that $\overrightarrow{\text{AB}}+\overrightarrow{\text{AC}}=\lambda\overrightarrow{\text{AD}}$Let $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are the position vectors of AB, BC and CA.
Now, the position vector of D is $\frac{\vec{\text{b}}+\vec{\text{c}}}2$. Then,
$\overrightarrow{\text{AB}}=\vec{\text{b}}-\vec{\text{a}}$
$\overrightarrow{\text{AC}}=\vec{\text{c}}-\vec{\text{a}}$
$\overrightarrow{\text{AD}}=\frac{\vec{\text{b}}+\vec{\text{c}}}2-\vec{\text{a}}$
Now, we have
$\overrightarrow{\text{AB}}+\overrightarrow{\text{AC}}=\lambda\overrightarrow{\text{AD}}$
$\Rightarrow\ \vec{\text{b}}-\vec{\text{a}}+\vec{\text{c}}-\vec{\text{a}}=\lambda\Big(\frac{\vec{\text{b}}+\vec{\text{c}}-\vec{\text{a}}}2\Big)$
$\Rightarrow\ \vec{\text{b}}+\vec{\text{c}}-2\vec{\text{a}}=\lambda\Big(\frac{\vec{\text{b}}+\vec{\text{c}}-2\vec{\text{a}}}2\Big)$
$\Rightarrow\ \lambda=2$
View full question & answer→Question 843 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 853 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 863 Marks
Show that the points A(1, 2, 7), B(2, 6, 3) and C(3, 10, -1) are collinear.
AnswerVertices A, B, C of a triangle are A (1, 2, 7), B(2, 6, 3) and C(3, 10, -1) respectively.
$\therefore\ \ \ \text{Position vector of point}\ \text{A}=\overrightarrow{\text{OA}}$ $=(1, 2, 7)=\hat{i}+2\hat{j}+7\hat{k}$
$\text{Position vector of point}\ \text{B}=\overrightarrow{\text{OB}}$ $=(2, 6, 3)=2\hat{i}+6\hat{j}+3\hat{k}$
$\text{Position vector of point}\ \text{C}=\overrightarrow{\text{OC}}$ $=(3, 10, -1)=3\hat{i}+10\hat{j}-\hat{k}$
Now $\ \ \overrightarrow{\text{AB}}$ = Position vector of point B - Position vector of point A
$=2\hat{i}+6\hat{j}+3\hat{k}-\big(\hat{i}+2\hat{j}+7\hat{k}\big)$
$=2\hat{i}+6\hat{j}+3\hat{k}-\hat{i}-2\hat{j}-7\hat{k}=\hat{i}+4\hat{j}-4\hat{k}\ \ \ \ \ \ \text{ ........(i)}$
And $\ \overrightarrow{\text{AC}}$ = Position vector of Point C - Position vector of point A
$=3\hat{i}+10\hat{j}-\hat{k}-(\hat{i}+2\hat{j}+7\hat{k})$
$=3\hat{i}+10\hat{j}-\hat{k}- \hat{i}-2\hat{j}-7\hat{k}$ $=2\hat{i}+8\hat{j}-8\hat{k}=2(\hat{i}+4\hat{j}-4\hat{k})\ \ \ \ \ ......\text{(ii)}$
$\Rightarrow\ \ \overrightarrow{\text{AC}}=2.\overrightarrow{\text{AB}}\ \ \ \big[\text{Using eq. (i)}\big]$
$\Rightarrow\ \text{Vectors}\ \overrightarrow{\text{AB}}\ \text{and}\ \overrightarrow{\text{AC}}\ \text{are collinear and parallel.}$ $\ \ \Big[\because\ \vec{a}=m\vec{b}\Big]$
Thus, Points A, B and C are collinear.
And also vectors $\overrightarrow{\text{AB}}\ \text{and}\ \overrightarrow{\text{AC}}$ have a common point A and hence can't be parallel.
View full question & answer→Question 873 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 883 Marks
Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are $(2\vec{\text{a}}+\vec{\text{b}})\ \text{and}\ (\vec{\text{a}}-3\vec{\text{b}})$ externally in the ratio 1 : 2. Also, show that P is the mid point of the line segment RQ.
AnswerIt is given that $\overrightarrow{\text{OP}}=2\vec{\text{a}}+\vec{\text{b}},\ \overrightarrow{\text{OQ}}=\vec{\text{a}}-3\vec{\text{b}}$It is given that point R divides a line segment joining two points P and Q externally in the ratio 1 : 2. Then, on using the section formula, we get:
$\overrightarrow{\text{OR}}=\frac{2\big(2\vec{\text{a}}+\vec{\text{b}}\big)-\big(\vec{\text{a}}-3\vec{\text{b}}\big)}{2-1}$ $\frac{4\vec{\text{a}}+2\vec{\text{b}}-\vec{\text{a}}+3\vec{\text{b}}}{1}=3\vec{\text{a}}+5\vec{\text{b}}$
Therefore, the position vector of point R is $3\vec{\text{a}}+5\vec{\text{b}}$
Position vector of the mid-point of RQ $=\frac{\overrightarrow{\text{OQ}}+\overrightarrow{\text{OR}}}{2}$
$=\frac{\big(\vec{\text{a}}-3\vec{\text{b}}\big)+\big(3\vec{\text{a}}+5\vec{\text{b}}\big)}{2}$
$2\vec{\text{a}}+\vec{\text{b}}$
$=\overrightarrow{\text{OP}}$
Hence, P is the mid-point of the line segment RQ.
View full question & answer→Question 893 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 903 Marks
Find the area of the parallelogram determinrd by the vectors:
$2\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k}}$ and $\hat{\text{i}}-\hat{\text{j}}$
AnswerLet:
$\vec{\text{a}}=2\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k}}$
$\vec{\text{b}}=\hat{\text{i}}-\hat{\text{j}}+0\hat{\text{k}}$
$\therefore\vec{\text{a}}\times\vec{\text{b}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\2&1&3\\1&-1&0 \end{vmatrix}$
$=(0+3)\hat{\text{i}}-(0-3)\hat{\text{j}}+(-2-1)\hat{\text{k}}$
$=3\hat{\text{i}}+3\hat{\text{j}}-3\hat{\text{k}}$
Area of the parallelogram $=\big|\vec{\text{a}}\times\vec{\text{b}}\big|$
$=\sqrt{3^2+3^2+3^2}$
$=\sqrt{27}$
$=3\sqrt{3}\text{ sq. units}$
View full question & answer→Question 913 Marks
ABCD is a quadrilateral. Find the sum of the vectors $\overrightarrow{\text{BA}},\overrightarrow{\text{BC}},\overrightarrow{\text{CD}}\text{ and }\overrightarrow{\text{DA}}$.
AnswerGiven: ABCD is a quadrilateral.
We need to find the sum of $\overrightarrow{\text{BA}}+\overrightarrow{\text{BC}}+\overrightarrow{\text{CD}}+\overrightarrow{\text{DA}}$.
Consider,
$\overrightarrow{\text{BA}}+\overrightarrow{\text{BC}}+\overrightarrow{\text{CD}}+\overrightarrow{\text{DA}}$
$=\Big(\overrightarrow{\text{BA}}+\overrightarrow{\text{DA}}\Big)+\Big(\overrightarrow{\text{BC}}+\overrightarrow{\text{CD}}\Big)$
$=\Big(\overrightarrow{\text{BD}}+2\overrightarrow{\text{DA}}\Big)+\overrightarrow{\text{BD}}$ $\Big[\because\ \overrightarrow{\text{BD}}+\overrightarrow{\text{DA}}=\overrightarrow{\text{BA}}\text{ and }\overrightarrow{\text{BC}}+\overrightarrow{\text{CD}}=\overrightarrow{\text{BD}}\Big]$
$=2\Big(\overrightarrow{\text{BD}}+\overrightarrow{\text{DA}}\Big)$
$=2\ \overrightarrow{\text{BA}}$
View full question & answer→Question 923 Marks
$\text{If} \ \vec{a}=2\hat{i}+2\hat{j}+3\hat{k},\ \ \vec{b}=-$ $\hat{i}+2\hat{j}+\hat{k}\ \text{and}\ \vec{c}=3\hat{i}+\hat{j}$ are such that $\vec{a}+\lambda\vec{b}$ is perpendicular to $\vec{c},$ then find the value of $\lambda.$
Answer$\text{Given:}\ \ \ \ \vec{a}=2\hat{i}+2\hat{j}+3\hat{k},\ \vec{b}=-$ $\hat{i}+2\hat{j}+\hat{k}\ \text{and}\ \vec{c}=3\hat{i}+\hat{j}$$\text{Now}\ \ \ \ \vec{a}+\lambda\vec{b}=2\hat{i}+2\hat{j}+3\hat{k}+\lambda(-\hat{i}+2\hat{j}+\hat{k})$ $=2\hat{i}+2\hat{j}+3\hat{k}-\lambda\hat{i}+2\lambda\hat{j}+\lambda\hat{k}$
$\Rightarrow\ \ \vec{a}+\lambda\vec{b}=(2-\lambda)\hat{i}+(2+2\lambda)\hat{j}+(3+\lambda)\hat{k}$
$\text{Again},\ \vec{c}=3\hat{i}+\hat{j}=3\hat{i}+\hat{j}+0\hat{k}$
$\text{Since},\ (\vec{a}+\lambda\vec{b})$ is perpendicular to $\vec{c},$ therefore, $\ \ (\vec{a}+\lambda\vec{b}).\vec{c}=0$
$\Rightarrow\ (2-\lambda)3+(2+2\lambda)1+(3+\lambda)0=0$
$\Rightarrow\ 6-3\lambda+2+2\lambda=0$ $\Rightarrow\ -\lambda+8=0$
$\Rightarrow\ -\lambda=-8\ \ \ \Rightarrow\ \lambda=8$
View full question & answer→Question 933 Marks
Find a unit vector in the direction of $\overrightarrow{\text{PQ}},$ where P and Q have co-ordinates (5, 0, 8) and (3, 3, 2), respectively.
AnswerSince, the e coordinates of P and Q are (5, 0, 8) and (3, 3, 2), respectively.
$\therefore\overrightarrow{\text{PQ}}=\overrightarrow{\text{OQ}}-\overrightarrow{\text{OP}}$
$=(3\hat{\text{i}}+3\hat{\text{j}}+2\hat{\text{k}})-(5\hat{\text{i}}+0\hat{\text{j}}+8\hat{\text{k}})$
$=-2\hat{\text{i}}+3\hat{\text{j}}-6\hat{\text{k}}$
View full question & answer→Question 943 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 953 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 963 Marks
Find the area of the parallelogram determinrd by the vectors:
$3\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}}$ and $\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}}$
AnswerLet:$\vec{\text{a}}=3\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}}$
$\vec{\text{b}}=1\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}}$
$\vec{\text{a}}\times\vec{\text{b}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\3&1&-2\\1&-3&4 \end{vmatrix}$
$=\hat{\text{i}}(4-6)-\hat{\text{j}}(12+2)+\hat{\text{k}}(-9-1)$
$=-2\hat{\text{i}}-14\hat{\text{j}}-10\hat{\text{k}}$
Area of the parallelogram $=\big|\vec{\text{a}}\times\vec{\text{b}}\big|$
$=\sqrt{(-2)^2+(-14)^2+(-10)^2}$
$=\sqrt{300}$
$=10\sqrt{3}\text{ sq. units}$
View full question & answer→Question 973 Marks
If $\vec{\text{a}}=4\hat{\text{i}}+3\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{b}}=\hat{\text{i}}-2\hat{\text{k}},$ then find $\big|2\hat{\text{b}}\times\vec{\text{a}}\big|.$
AnswerGiven:
$\vec{\text{a}}=4\hat{\text{i}}+3\hat{\text{j}}+\hat{\text{k}}$
$2\vec{\text{b}}=2\hat{\text{i}}+0\hat{\text{j}}+4\hat{\text{k}}$
$2\vec{\text{b}}\times\vec{\text{a}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\2&0&-4\\4&3&1 \end{vmatrix}$
$=(0+12)\hat{\text{i}}-(2+16)\hat{\text{j}}+(6-0)\hat{\text{k}}$
$=12\hat{\text{i}}-18\hat{\text{j}}+6\hat{\text{k}}$
$\Rightarrow\big|2\vec{\text{b}}\times\vec{\text{a}}\big|=\sqrt{12^2+(-18^2)+6^2}$
$=\sqrt{504}$
View full question & answer→Question 983 Marks
Find the area of the triangle formed by O, A, B when $\overrightarrow{\text{OA}}=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}},\overrightarrow{\text{OB}}=-3\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}.$
AnswerGiven: $\overrightarrow{\text{OA}}=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$ $\overrightarrow{\text{OB}}=-3\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}$ $\overrightarrow{\text{OA}}\times\overrightarrow{\text{OB}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\1&2&3\\-3&-2&1 \end{vmatrix}$ $=8\hat{\text{i}}-10\hat{\text{j}}+4\hat{\text{k}}$ $\big|\overrightarrow{\text{OA}}\times\overrightarrow{\text{OB}}\big|=\sqrt{64+100+16}$ $=\sqrt{180}$ $=6\sqrt{5}$ Area of the triangle $=\frac{1}{2}\big|\overrightarrow{\text{OA}}\times\overrightarrow{\text{OB}}\big|$ $=\frac{1}{2}(6\sqrt{5})$$=3\sqrt{5}\text{ sq. units}$
View full question & answer→Question 993 Marks
ABCDE is a pentagon, prove that,$\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}+\overrightarrow{\text{CD}}+\overrightarrow{\text{DE}}+\overrightarrow{\text{EA}}=\vec0$
AnswerGiven ABCDE is a pentagon.
$\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}+\overrightarrow{\text{CD}}+\overrightarrow{\text{DE}}+\overrightarrow{\text{EA}}$
$=\Big(\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}\Big)+\overrightarrow{\text{CD}}+\overrightarrow{\text{DE}}+\overrightarrow{\text{EA}}=\vec0$
$=\overrightarrow{\text{AC}}+\overrightarrow{\text{CD}}+\overrightarrow{\text{DE}}+\overrightarrow{\text{EA}}$ $\Big[$Using triangle law in $\triangle\text{ABC},\ \overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}=\overrightarrow{\text{AC}}\Big]$
$=\Big(\overrightarrow{\text{AC}}+\overrightarrow{\text{CD}}\Big)+\overrightarrow{\text{DE}}+\overrightarrow{\text{EA}}$
$=\overrightarrow{\text{AD}}+\overrightarrow{\text{DE}}+\overrightarrow{\text{EA}}$ $\Big[$Using triangle law in $\triangle\text{ACD},\ \overrightarrow{\text{AC}}+\overrightarrow{\text{CD}}=\overrightarrow{\text{AD}}\Big]$
$=\overrightarrow{\text{AD}}+\overrightarrow{\text{DA}}$
$=\overrightarrow{\text{AD}}-\Big(-\overrightarrow{\text{AD}}\Big)$
$=\vec0$
$\therefore\ \overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}+\overrightarrow{\text{CD}}+\overrightarrow{\text{DE}}+\overrightarrow{\text{EA}}=\vec0$
View full question & answer→Question 1003 Marks
Find the angle between two vectors $\vec{\text{a}}$ and $\vec{\text{b},}$ if $\big|\vec{\text{a}}\times\vec{\text{b}}\big|=\vec{\text{a}}.\vec{\text{b}}.$
AnswerLet $\theta$ be the angle between $\vec{\text{a}}$ and $\vec{\text{b}. }$
Given:
$\big|\vec{\text{a}}\times\vec{\text{b}}\big|=\vec{\text{a}}.\vec{\text{b}}$
$\Rightarrow|\vec{\text{a}}|\big|\vec{\text{b}}\big|\sin\theta=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta$
$\Rightarrow\sin\theta=\cos\theta$
$\Rightarrow\tan\theta=1$
$\Rightarrow\theta=\frac{\pi}{4}$
View full question & answer→