5 Marks Questions

Question 1515 Marks

A vector $\vec{\text{r}}$ is inclined at equal angles to the three axes. If the magnitude of $\vec{\text{r}}$ is $2\sqrt{3}$ units, find $\vec{\text{r}}.$

Answer

We have, $|\vec{\text{r}}|=2\sqrt{3}$
Since $\vec{\text{r}}$ is equally inclined to the three axes, $\vec{\text{r}}$ so direction cosines of the unit vector $\vec{\text{r}}$ will be same i.e., $\text{l}=\text{m}=\text{n}.$
We know that,
$\text{l}^2+\text{m}^2+\text{n}^2=1$
$\Rightarrow\text{l}^2+\text{l}^2+\text{l}^2=1$
$\Rightarrow\text{l}^2=\frac{1}{3}$
$\Rightarrow\text{l}=\pm\Big(\frac{1}{\sqrt{3}}\Big)$
So, $\hat{\text{r}}=\pm\frac{1}{\sqrt{3}}\hat{\text{i}}\pm\frac{1}{\sqrt{3}}\hat{\text{j}}\pm\frac{1}{\sqrt{3}}\hat{\text{k}}$
$\therefore\vec{\text{r}}=\hat{\text{r}}|\hat{\text{r}}|\ \Big[\because\hat{\text{r}}=\frac{\vec{\text{r}}}{|\vec{\text{r}}|}\Big]$
$=\Big[\pm\frac{1}{\sqrt{3}}\hat{\text{i}}\pm\frac{1}{\sqrt{3}}\hat{\text{j}}\pm\frac{1}{\sqrt{3}}\hat{\text{k}}\Big]2\sqrt{3}$ $\big[\because|\text{r}|=2\sqrt{3}\big]$
$=\pm2\hat{\text{i}}+\pm2\hat{\text{j}}+\pm2\hat{\text{k}}$
$=\pm2(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})$

View full question & answer→

Question 1525 Marks

Using vectors, prove that the parallelogram on the same base and between the same parallels are equal in area.

Answer

Let ABCD and ABFE are parallelograms on the same base AB and between the same parallel line AB and DF.

Let $\overrightarrow{\text{AB}}=\vec{\text{a}}$ and $\overrightarrow{\text{AD}}=\vec{\text{b}}$
$\therefore$ Area of parallelogram $\overrightarrow{\text{ABCD}}=\vec{\text{a}}\times\vec{\text{b}}$
Now, area of parallelogram of $\overrightarrow{\text{ABFE}}=\overrightarrow{\text{AB}}\times\overrightarrow{\text{AE}}$
$=\overrightarrow{\text{AB}}\times(\overrightarrow{\text{AD}}+\overrightarrow{\text{DE}})$
$=\overrightarrow{\text{AB}}\times(\vec{\text{b}}+\text{k}\vec{\text{a}})$ $[\text{let}\overrightarrow{\text{ DE}}=\text{k}\vec{\text{a}},\text{where k is a scalar}]$
$=\vec{\text{a}}\times(\vec{\text{b}}+\text{k}\vec{\text{a}})$
$=(\vec{\text{a}}\times\vec{\text{b}})+(\vec{\text{a}}\times\text{k}\vec{\text{a}})$
$=(\vec{\text{a}}\times\vec{\text{b}})+\text{k}(\vec{\text{a}}\times\vec{\text{a}})$
$=(\vec{\text{a}}\times\vec{\text{b}})\ [\because\vec{\text{a}}\times\vec{\text{a}}=0]$
Area of parallelogram ABCD.
Hence proved.

View full question & answer→

Question 1535 Marks

Prove using vector: the quadrilateral obtained by joining mid-points of adjacent sides of a rectangle is a rhombus.

Answer

ABCD is a rectangle. Let P, Q, R and S be the mid-points of the sides AB, BC, CD and DA, respectively.Now,
$\vec{\text{PQ}} = \vec{\text{PB}} + \vec{\text{BQ}} = \frac{1}{2}\vec{\text{AB}} + \frac{1}{2}\vec{\text{BC}}$
$=\frac{1}{2}\big(\vec{\text{AB}}+\vec{\text{BC}}\big)=\frac{1}{2}\vec{\text{AC}}\dots(1)$
$\vec{\text{SR}} = \vec{\text{SD}} + \vec{\text{DR}} = \frac{1}{2}\vec{\text{AD}} + \frac{1}{2}\vec{\text{DC}}$
$=\frac{1}{2}\big(\vec{\text{AD}}+\vec{\text{DC}}\big)=\frac{1}{2}\vec{\text{AC}}\dots(2)$
From (1) and (2), we have
$\vec{\text{PQ}} = \vec{\text{SR}}$
So, the sides PQ and SR are equal and parallel. Thus, PQRS is a parallelogram.
Now,
$\big|\vec{\text{PQ}}\big|^{2}=\vec{\text{PQ}}.\vec{\text{PQ}}$
$\Rightarrow \big|\text{PQ}\big|^{2}=\big(\vec{\text{PB}} + \vec{\text{BQ}}\big).\big(\vec{\text{PB}} + \vec{\text{BQ}}\big)$
$\Rightarrow \big|\vec{\text{PQ}}\big|^2= \big|\vec{\text{PB}}\big|^{2} + 2\vec{\text{PB}}.\vec{\text{BQ}} + \big|\vec{\text{BQ}}\big|^{2}$
$\Rightarrow \big|\vec{\text{PQ}}\big|^2 = \big|\vec{\text{PB}}\big|^{2} + 0 + \big|\vec{\text{BQ}}\big|^2 $ $\big(\vec{\text{PB}} \perp \vec{\text{BQ}}\big)$
$\Rightarrow \big|\vec{\text{PQ}}\big|^{2} = \big|\vec{\text{PB}}\big|^{2} + \big|\vec{\text{BQ}}\big|^{2}\dots(3)$
Also,
$\big|\vec{\text{PS}}\big|^{2} = \vec{\text{PS}}.\vec{\text{PS}}$
$\Rightarrow\big|\vec{\text{PS}}\big|^{2} = \big(\vec{\text{PA}} + \vec{\text{AS}}\big).\big(\vec{\text{PA}} + \vec{\text{AS}}\big)$
$\Rightarrow \big|\vec{\text{PS}}\big|^{2} = \big|\vec{\text{PA}}\big|^{2}+ 2\vec{\text{PA}}.\vec{\text{AS}} + \big|\vec{\text{AS}}\big|^{2}$
$\Rightarrow \big|\vec{\text{PS}}\big|^{2} = \big|\vec{\text{PB}}\big|^{2} + 0 + \big|\vec{\text{BQ}}\big|^{2}$ $\big(\vec{\text{PA}} \perp \vec{\text{AS}}\big)$
$\Rightarrow \big|\vec{\text{PS}}\big|^{2} = \big|\vec{\text{PB}}\big|^{2} + \big|\vec{\text{BQ}}\big|^{2}\dots(4)$
From (3) and (4), we have
$\big|\vec{\text{PQ}}\big|^{2} = \big|\vec{\text{PS}}\big|^{2}$
$\Rightarrow \big|\vec{\text{PQ}}\big| = \big|\vec{\text{PS}}\big|$
So, the adjacent sides of the parallelogram are equal. Hence, PQRS is a rhombus.

View full question & answer→

Question 1545 Marks

Show that each of the given three vectors is a unit vector:$\frac{1}{7}(2\hat{i}+3\hat{j}+6\hat{k}),\ \frac{1}{7}(3\hat{i}-6\hat{j}+2\hat{k}),\ \frac{1}{7}(6\hat{ i}+2\hat{j}-3\hat{k})$
Also, show that they are mutually perpendicular to each other.

Answer

$\text{Let}\ \ \vec{a}=\frac{1}{7}\Big(2\hat{i}+3\hat{j}+6\hat{k}\Big)$ $=\frac{2} {7}\hat{i}+\frac{3}{7}\hat{j}+\frac{6}{7}\hat{k}\ \ \ \ \ .......\text{(i)}$$\vec{b}=\frac{1}{7}\Big(3\hat{i}-6\hat{j}+2\hat{k}\Big)$ $=\frac{3} {7}\hat{i}-\frac{6}{7}\hat{j}+\frac{2}{7}\hat{k}\ \ \ \ \ .......\text{(ii)}$
$\vec{c}=\frac{1}{7}\Big(6\hat{i}+2\hat{j}-3\hat{k}\Big)$ $=\frac{6} {7}\hat{i}+\frac{2}{7}\hat{j}-\frac{3}{7}\hat{k}\ \ .......\text{(iii)}$
$\Rightarrow\ \ \ \Big|\vec{a}\Big|=\sqrt{\Big(\frac{2}{7}\Big)^2+\Big(\frac{3}{7}\Big)^2+\Big(\frac{6}{7}\Big)^2}$ $=\sqrt{\frac{4}{49}+\frac{9}{49}+\frac{36}{49}}=\sqrt{\frac{49}{49}}=\sqrt{1}=1$$\Big|\vec{b}\Big|=\sqrt{\Big(\frac{3}{7}\Big)^2+\Big(\frac{-6}{7}\Big)^2+\Big(\frac{2}{7}\Big)^2}$ $=\sqrt{\frac{9}{49}+\frac{36}{49}+\frac{4}{49}}=\sqrt{\frac{49}{49}}=\sqrt{1}=1$
$\big|\vec{c}\big|=\sqrt{\Big(\frac{6}{7}\Big)^2+\Big(\frac{2}{7}\Big)^2+\Big(\frac{-3}{7}\Big)^2}$ $=\sqrt{\frac{36}{49}+\frac{4}{49}+\frac{9}{49}}=\sqrt{\frac{49}{49}}=\sqrt{1}=1$
$\therefore$ Each of the three given vectors $\vec{a},\vec{b},\vec{c}$ is a unit vector.From eq. (i) and (ii),
$\vec{a}.\vec{b}=\Big(\frac{2}{7}\Big).\Big(\frac{3}{7}\Big)+\Big(\frac{3}{7}\Big).\Big(\frac{-6}{7}\Big)+\Big(\frac{6}{7}\Big).\Big(\frac{2}{7}\Big)$ $\Big[\because\vec{a}.\vec{b}=a_1b_1+a_2b_2+a_3b_3\Big]$
$=\frac{6}{49}-\frac{18}{49}+\frac{12}{49}=\frac{6-18+12}{49}=\frac{0}{49}=0$
$\therefore\ \ \vec{a}\ \text{and}\ \vec{b}$ are perpendicular to each other. From eq. (ii) and eq. (iii),$\vec{b}.\vec{c}=\Big(\frac{3}{7}\Big).\Big(\frac{6}{7}\Big)+\Big(\frac{-6}{7}\Big).\Big(\frac{2}{7}\Big).\Big(\frac{2}{7}\Big).\Big(\frac{-3}{7}\Big)$$\Big[\because\vec{a}.\vec{b}=a_1b_1+a_2b_2+a_3b_3\Big]$
$=\frac{18}{49}-\frac{12}{49}-\frac{6}{49}=\frac{18-12-6}{49}=\frac{0}{49}=0 $
$\therefore\ \ \vec{a}\ \text{and}\ \vec{b}$ are perpendicular to each other. From eq. (i) and (iii),$\vec{a}.\vec{c}=\Big(\frac{2}{7}\Big).\Big(\frac{6}{7}\Big)+\Big(\frac{3}{7}\Big).\Big(\frac{2}{7}\Big)+\Big(\frac{6}{7}\Big).\Big(\frac{-3}{7}\Big)$ $\Big[\because\vec{a}.\vec{b}=a_1b_1+a_2b_2+a_3b_3\Big]$
$=\frac{12}{49}+\frac{6}{49}-\frac{18}{49}=\frac{12+6-18}{49}=\frac{0}{49}=0 $
$\therefore\ \ \vec{a}\ \text{and}\ \vec{b}$ are perpendicular to each other. Hence, $\vec{a},\vec{b},\vec{c}$ are mutually perpendicular vectors.

View full question & answer→

Question 1555 Marks

$\text{If}\ \vec{\text{a},}\ \vec{\text{b}},\ \vec{\text{c}}$ are mutually perpendicular vectors of equal magnitudes, show that the vector $\vec{\text{a}}+ \vec{\text{b}}+ \vec{\text{c}}$ is equally inclined to $\vec{\text{a},}\ \vec{\text{b}},\text{and}\ \vec{\text{c}}.$

Answer

Since $\vec{\text{a},}\ \vec{\text{b}},\text{and}\ \vec{\text{c}}$ are mutually perpendicular vectors, we have $\vec{\text{a}}\cdot\vec{\text{b}}=\vec{\text{b}}\cdot\vec{\text{c}}=\vec{\text{c}}\cdot\vec{\text{a}}=0.$ It is given that: $\big|\vec{\text{a}}\big|=\Big|\vec{\text{b}}\Big|=\big|\vec{\text{c}}\big|$ Let vector $\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}$ be inclined to $\vec{\text{a}},\vec{\text{b}},\ \text{and}\ \vec{\text{c}}$ at angles $\theta_{1},\ \theta_{2},\ \text{and}\ \theta_{3}$ respectively. Then, we have: $\cos\theta_{1}=\frac{\big(\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big)\cdot\vec{\text{a}}}{\big|\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big|\big|\vec{\text{a}}\big|}=\frac{\vec{\text{a}}\cdot\vec{\text{a}}+\vec{\text{b}}\cdot\vec{\text{a}}+\vec{\text{c}}\cdot\vec{\text{a}}}{\big|\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big|\big|\vec{\text{a}}\big|}$ $=\frac{\big|\vec{\text{a}}\big|^2}{\Big|\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\Big|\big|\vec{\text{a}}\big|}\ \ \Big[\vec{\text{b}}\cdot\vec{\text{a}}=\vec{\text{c}}\cdot\vec{\text{a}}=0\Big]$ $=\frac{\big|\vec{\text{a}}\big|}{\Big|\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\Big|}$ $\cos\theta_{2}=\frac{\Big(\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\Big)\cdot\vec{\text{b}}}{\Big|\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\Big|\Big|\vec{\text{b}}\Big|}=\frac{\vec{\text{a}}\cdot\vec{\text{b}}+\vec{\text{b}}\cdot\vec{\text{b}}+\vec{\text{c}}\cdot\vec{\text{b}}}{\Big|\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\Big|\cdot\Big|\vec{\text{b}}\Big|}$ $=\frac{\Big|\vec{\text{b}}\Big|^2}{\Big|\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\Big|\cdot\Big|\vec{\text{b}}\Big|}\ \ \Big[\vec{\text{a}}\cdot\vec{\text{b}}=\vec{\text{c}}\cdot\vec{\text{b}}=0\Big]$ $=\frac{\Big|\vec{\text{b}}\Big|}{\Big|\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\Big|}$ $ \cos\theta_{3}=\frac{\Big(\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\Big)\cdot\vec{\text{c}}}{\Big|\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\Big|\big|\vec{\text{c}}\big|}=\frac{\vec{\text{a}}\cdot\vec{\text{c}}+\vec{\text{b}}\cdot\vec{\text{c}}+\vec{\text{c}}\cdot\vec{\text{c}}}{\Big|\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\Big|\big|\vec{\text{c}}\big|}$$=\frac{\big|\vec{\text{c}}\big|^2}{\Big|\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\Big|\big|\vec{\text{c}}\big|}\ \ \Big[\vec{\text{a}}\cdot\vec{\text{c}}=\vec{\text{b}}\cdot\vec{\text{c}}=0\Big]$
$=\frac{\big|\vec{\text{c}}\big|}{\Big|\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\Big|}$
$\text{Now, as}|\vec{\text{a}}|=\Big|\vec{\text{b}}\Big|=|\vec{\text{c}}|,\ \cos\theta_1=\cos\theta_2=\cos\theta_3.$ $\therefore\ \theta_1=\theta_2=\ \theta_3$ Hence, the vector $\Big(\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\Big)$ is equally inclined to $\vec{\text{a}},\vec{\text{b}}\ \text{and}\ \vec{\text{c}}.$

View full question & answer→

Question 1565 Marks

Express the vector $\vec{\text{a}}=5\hat{\text{i}}-2\hat{\text{j}}+5\hat{\text{k}}$ as the sum of two vectors such that one is parallel to the vector $\vec{\text{b}}=3\hat{\text{i}}+\hat{\text{k}}$ and other is perpendicular to $\vec{\text{b}}.$

Answer

Given that $\vec{\text{a}}=5\hat{\text{i}}-2\hat{\text{j}}+5\hat{\text{k}}$ and $\vec{\text{b}}=3\hat{\text{i}}+\hat{\text{k}}$
Let $\vec{\text{x}}$ and $\vec{\text{y}} $ be such that
$\vec{\text{a}} =\vec{\text{x}} +\vec{\text{y}} $
$\Rightarrow\vec{\text{y}} =\vec{\text{a}} -\vec{\text{x}} \dots(1)$
Since $\vec{\text{x}}$ is parallei to $\vec{\text{b}},$
$\Rightarrow\vec{\text{x}}=\text{t}\vec{\text{b}}$ (t is constant)
$\Rightarrow\vec{\text{x}}=\text{t}\big(3\hat{\text{i}}+\vec{\text{k}}\big)=3\text{t}\hat{\text{i}}+\text{t}\hat{\text{k}}$
Substituting the values of $\vec{\text{x}}$ and $\vec{\text{a}}$ in (1), we get
$\vec{\text{y}}=5\hat{\text{i}}-2\hat{\text{j}}+5\hat{\text{k}}-\big(3\text{t}\hat{\text{i}}+\text{t}\hat{\text{k}}\big)=(5-3\text{t})\hat{\text{i}}-2\hat{\text{j}}+(5-\text{t})\hat{\text{k}}\dots(2)$
Since $\vec{\text{y}} $ is perpendicular to $\vec{\text{b}},$
$\vec{\text{y}}.\vec{\text{b}}=0$
$\Rightarrow\big[(5-3\text{t})\hat{\text{i}}-2\hat{\text{j}}+(5-\text{t})\hat{\text{k}}\big].\big(3\hat{\text{i}}+\hat{\text{k}}\big)=0$
$\Rightarrow3(5-3\text{t})+0+(5-\text{t})=0$
$\Rightarrow15-9\text{t}+5-\text{t}=0$
$\Rightarrow20-10\text{t}=0$
$\Rightarrow\text{t}=2$
From (1) and (2),we get
$\vec{\text{x}}=6\hat{\text{i}}+2\hat{\text{k}}$
$\vec{\text{y}}=-\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}$

View full question & answer→

Question 1575 Marks

Show that area of the parallelogram whose diagonals ate given by $\vec{\text{a}}$ and $\vec{\text{b}}$ is $\frac{|\vec{\text{a}}\times\vec{\text{b}}|}{2}.$ Also, find the area of the parallelogram, whose diagonals are $2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$ and $\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}}.$

Answer

Let ABCD be a parallelogram swuch that
$\overrightarrow{\text{AB}}=\vec{\text{p}},\overrightarrow{\text{AD}}=\vec{\text{q}}\Rightarrow\overrightarrow{\text{BC}}=\vec{\text{q}}$
$\therefore\overrightarrow{\text{AC}}=\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}=\vec{\text{p}}+\vec{\text{q}}=\vec{\text{a}}\ ....(\text{i})$
And $\therefore\overrightarrow{\text{BD}}=\overrightarrow{\text{BA}}+\overrightarrow{\text{AD}}=-\vec{\text{p}}+\vec{\text{q}}=\vec{\text{b}}\ ....(\text{ii})$
Adding (i) and (ii), we get
$\vec{\text{a}}+\vec{\text{b}}=2\vec{\text{q}}$
Subbtracting (ii) from Eq. (i), we get
$\vec{\text{a}}-\vec{\text{b}}=2\vec{\text{p}}$
$\Rightarrow\vec{\text{p}}=\frac{1}{2}(\vec{\text{a}}-\vec{\text{b}})$
Now, $\vec{\text{p}}\times\vec{\text{q}}=\frac{1}{4}(\vec{\text{a}}-\vec{\text{b}})\times(\vec{\text{a}}+\vec{\text{b}})$
$=\frac{1}{4}(\vec{\text{a}}\times\vec{\text{a}}+\vec{\text{a}}\times\vec{\text{b}}-\vec{\text{b}}\times\vec{\text{a}}-\vec{\text{b}}\times\vec{\text{b}})$
$=\frac{1}{4}[\vec{\text{a}}\times\vec{\text{b}}+\vec{\text{a}}\times\vec{\text{b}}]$
$=\frac{1}{2}(\vec{\text{a}}\times\vec{\text{b}})$
So area of a parallelogram $\text{ABCD}=|\vec{\text{p}}\times\vec{\text{q}}|=\frac{1}{2}|\vec{\text{a}}\times\vec{\text{b}}|$
So, area of a parallelogram, whose diagonals are $2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$ and $\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}}$
$=\frac{1}{2}|(2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})\times(\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}})|$
$=\frac{1}{2}\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}} \\2&-1&1\\1&3&-1 \end{vmatrix}$
$=\frac{1}{2}|[\vec{\text{i}}(1-3)-\vec{\text{j}}(-2-1)+\vec{\text{k}}(6+1)]|$
$=\frac{1}{2}|-2\vec{\text{i}}+3\vec{\text{j}}+7\vec{\text{k}}|$
$=\frac{1}{2}\sqrt{4+9+49}$
$=\frac{1}{2}\sqrt{62}\text{ sq. units}$

View full question & answer→

Question 1585 Marks

Express $2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}$ as the sum of a vector parallel and a vector perpendicular to $2\hat{\text{i}}+4\hat{\text{j}}-2\hat{\text{k}}.$

Answer

Let $\big(2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}\big)=\vec{\text{a}}+\vec{\text{b}}\dots(1)$
such that $\vec{\text{a}}$ is a vector parallel to vector $\big(2\hat{\text{i}}+4\hat{\text{j}}-2\hat{\text{k}}\big)$ and $\vec{\text{b}}$ is a vector perpendicular to the vector $\big(2\hat{\text{i}}+4\hat{\text{j}}-2\hat{\text{k}}\big).$
since, $\vec{\text{a}}$ is parallel to $\big(2\hat{\text{i}}+4\hat{\text{j}}-2\hat{\text{k}}\big)$
$\vec{\text{a}}=\lambda\big(2\hat{\text{i}}+4\hat{\text{j}}-2\hat{\text{k}}\big)$
$\vec{\text{a}}=2\lambda\hat{\text{i}}+4\lambda\hat{\text{j}}-2\lambda\hat{\text{k}}\dots(2)$
Put value of $\vec{\text{a}}$ in equation (1),
$\big(2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}\big)=\big(2\lambda\hat{\text{i}}+4\lambda\hat{\text{j}}-2\lambda\hat{\text{k}}\big)+\vec{\text{b}}$
$\vec{\text{b}}=2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}-2\lambda\hat{\text{j}}-4\lambda\hat{\text{j}}+2\lambda\hat{\text{k}}$
$\vec{\text{b}}=(2-2\lambda)\hat{\text{i}}+(-1-4\lambda)\hat{\text{j}}+(3+2\lambda)\hat{\text{k}}$
$\vec{\text{b}}$ is a vector perpendicuar to the vector $\big(2\hat{\text{i}}+4\hat{\text{j}}-2\hat{\text{k}}\big),$ then
$\vec{\text{b}}.\big(2\hat{\text{i}}+4\hat{\text{j}}-2\hat{\text{k}}\big)=0$
$\big[(2-2\lambda)\hat{\text{i}}+(-1-4\lambda)\hat{\text{j}}+(3+2\lambda)\hat{\text{k}}\big]\big(2\hat{\text{i}}+4\hat{\text{j}}-2\hat{\text{k}}\big)=0$
$(2-2\lambda)(2)+(-1-4\lambda)(4)+(3+2\lambda)(-2)=0$
$4-4\lambda-4-16\lambda-6-4\lambda=0$
$-6-24\lambda=0$
$-24\lambda=6$
$\lambda=-\frac{1}{4}$
Put $\lambda$ in equation (2),
$\vec{\text{a}}=2\lambda\hat{\text{i}}+4\lambda\hat{\text{j}}-2\lambda\hat{\text{k}}$
$=2\Big(-\frac{1}{4}\Big)\hat{\text{i}}+4\Big(-\frac{1}{4}\Big)\hat{\text{j}}-2\Big(-\frac{1}{4}\Big)\hat{\text{k}}$
$\vec{\text{a}}=-\frac{1}2{}\hat{\text{i}}-\hat{\text{j}}+\frac{1}2{}\hat{\text{k}}$
Put the value of $\vec{\text{a}}$ in equation (1),
$\big(2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}\big)=\Big(-\frac{1}{2}\hat{\text{i}}-\hat{\text{j}}+\frac{1}{2}\hat{\text{k}}\Big)+\vec{\text{b}}$
$\vec{\text{b}}=2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}+\frac{1}{2}\hat{\text{i}}+\hat{\text{j}}-\frac{1}{2}\hat{\text{k}}$
$=\frac{4\hat{\text{i}}-2\hat{\text{j}}+6\hat{\text{k}}+\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}}{2}$
$=\frac{5\hat{\text{i}}+5\hat{\text{k}}}{2}$
$\vec{\text{b}}=\frac{5}{2}\big(\hat{\text{i}}+\hat{\text{k}}\big)$
$\big(2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}\big)=\Big(-\frac{1}{2}\hat{\text{i}}-\hat{\text{j}}+\frac{1}{2}\hat{\text{k}}\Big)+\frac{5}{2}\big(\hat{\text{i}}+\hat{\text{k}}\big)$

View full question & answer→

Question 1595 Marks

If $\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}=0,$ then show that $\vec{\text{a}}\times\vec{\text{b}}=\vec{\text{b}}\times\vec{\text{c}}=\vec{\text{c}}\times\vec{\text{a}}.$ Interpret the result geometrically?

Answer

Since, $\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}=0$
$\vec{\text{b}}=-\vec{\text{c}}-\vec{\text{a}}$
Now, $\vec{\text{a}}\times\vec{\text{b}}=\vec{\text{a}}\times(\vec{\text{c}}\times\vec{\text{a}})$
$=\vec{\text{a}}\times(-\vec{\text{c}})=\vec{\text{a}}\times(\vec{\text{c}}-\vec{\text{a}})=-\vec{\text{a}}\times\vec{\text{c}}$
$\Rightarrow\vec{\text{a}}\times\vec{\text{b}}=\vec{\text{c}}\times\vec{\text{a}}\ .....(\text{i})$
Also, $\vec{\text{b}}\times\vec{\text{c}}=(-\vec{\text{c}}-\vec{\text{a}})\times\vec{\text{c}}$
$=(-\vec{\text{c}}\times\vec{\text{c}})+(-\vec{\text{a}}\times\vec{\text{c}})=-\vec{\text{a}}\times\vec{\text{c}}$
$\Rightarrow\vec{\text{b}}\times\vec{\text{c}}=\vec{\text{c}}\times\vec{\text{a}}\ .....(\text{ii})$
From equations (i) and (ii), we have
$\vec{\text{a}}\times\vec{\text{b}}=\vec{\text{b}}\times\vec{\text{c}}=\vec{\text{c}}\times\vec{\text{a}}$
Geometrical interpretation of the result
If ABCD is a parallelogram such that $\overrightarrow{\text{AB}}=\vec{\text{a}}$ and $\overrightarrow{\text{AD}}=\vec{\text{b}}$ and these adjacent sides are making angle $\theta$ between each orther, then
Area of parallelogram $\text{ABCD}=|\vec{\text{a}}||\vec{\text{b}}||\sin\theta|=|\vec{\text{a}}\times\vec{\text{b}}|$

Since, parallelograms on the same base and between the same parallels are equal in area, so we have,
$|\vec{\text{a}}\times\vec{\text{b}}|=|\vec{\text{b}}\times\vec{\text{c}}|=|\vec{\text{b}}\times\vec{\text{a}}|$
This also implies that $\vec{\text{a}}\times\vec{\text{b}}=\vec{\text{a}}\times\vec{\text{c}}=\vec{\text{b}}\times\vec{\text{c}}$
So, area of the parallelograms formed by making any two sides represented by $\vec{\text{a}},\vec{\text{b}}$ and $\vec{\text{c}}$ as adjacent sides are equal.

View full question & answer→

MATHS 5 Marks Questions