Question 13 Marks
Find the projection of $\vec{b}+\vec{c}$ on $\vec{a}$ where $\vec{a}=\hat{i}+2\hat{j}+\hat{k},\text{ }\vec{b}=\hat{i}+3\hat{j}+\hat{k}$ and $\vec{c}=\hat{i}+\hat{k}.$
Answer$\vec{\text{b}}+\vec{\text{c}}=2\hat{\text{i}}+3\hat{\text{j}}+2\hat{\text{k}}$ Projection p = $\frac{(2\hat{\text{i}}+3\hat{\text{j}}+2\hat{\text{k}})\cdot{(\hat{\text{i}}+2\hat{\text{j}}}+\hat{\text{k}})}{|\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k|}}}$= $\frac{2 + 6 + 2}{\sqrt{6}}=\frac{10}{\sqrt{6}}\cdot\text{OR }\frac{5\sqrt{6}} {3} $.
View full question & answer→Question 23 Marks
Find the value of $\lambda$ which makes the vectors $\vec{a},\vec{b},\vec{c}$ coplanar, where $\vec{a}=-4\hat{\text{i}}-6\hat{\text{j}}-2\hat{\text{k}},\text{ }\vec{b}=-\hat{\text{i}}+4\hat{\text{j}}+3\hat{\text{k}}$ and $\vec{c}=-8\hat{\text{i}}-\hat{\text{j}}+\lambda\hat{\text{k}.}$
AnswerFor coplanarity $[\vec{a},\vec{b},\vec{c}]=0$$\therefore\begin{vmatrix} -4 & -6 & -2 \\ -1 & 4 & 3 \\ -8 & -1 & \lambda \end{vmatrix}=0$
$\Rightarrow$ – 8 (–18 + 8) + (–12 – 2) + $\lambda$ (–16 – 6) = 0
$\Rightarrow$ 80 – 14 – 22 $\lambda$ = 0 $\Rightarrow$ $\lambda$ = 3.
View full question & answer→Question 33 Marks
Find the angle between the vectors $\vec{\text{a}}+\vec{\text{b}}\text{ and }\vec{\text{a}}-\vec{\text{b}}\text{ if }\text{ }\vec{\text{a}}=2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}\text{ and }\vec{\text{b}}\text{ }=3\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}}.$
AnswerLet = $\vec{\text{p}}=\vec{\text{a}}+\vec{\text{b}}=\Big(5\hat{\text{i}}+\hat{\text{k}}\Big),\text{ }\vec{\text{p}}=\vec{\text{a}}+\vec{\text{b}}=\Big(-\hat{\text{i}}+2\hat{\text{j}}+5\hat{\text{k}}\Big)$
$\text{Using }\text{ }\cos\theta=\frac{\vec{\text{p}}\cdot\vec{\text{q}}}{|\vec{\text{p}}||\vec{\text{q}}|},\text{we get}$
$\cos\theta=0\text{ }\Rightarrow\text{ }\theta=\frac{\pi}{2}.$
View full question & answer→Question 43 Marks
Using vectors, prove that in a $\Delta$ ABC,$\frac{\text{a}}{\sin\text{A}}=\frac{\text{b}}{\sin\text{B}}=\frac{\text{c}}{\sin\text{C}}$
Where a, b and c are lengths of the sides opposite, respectively, to the angles A, B and C of $\Delta$ ABC.
Answer
Let in $\Delta$ ABC, BC = $\vec{\text{a}},\text{ }\text{ CA}=\vec{\text{b}}\text{ and }\text{ }\vec{\text{AB}}=\vec{\text{c}}$
$\therefore\text{ }\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}=\vec{\text{o}}$
$\vec{\text{a}}\times\vec{\text{a}}+\vec{\text{a}}\times\vec{\text{b}}+\vec{\text{a}}\times\vec{\text{c}}=\vec{\text{o}}\Rightarrow\vec{\text{a}}\times\vec{\text{b}}=\vec{\text{c}}\times\vec{\text{a}}............\text{(i)}$
and $\vec{\text{a}}\times\vec{\text{b}}+\vec{\text{b}}\times\vec{\text{b}}+\vec{\text{c}}\times\vec{\text{b}}=\vec{\text{o}}\Rightarrow\vec{\text{a}}\times\vec{\text{b}}=\vec{\text{b}}\times\vec{\text{c}}..........{\text{(ii)}}$
$\therefore\text{ }\vec{\text{a}}\times\vec{\text{b}}=\vec{\text{b}}\times\vec{\text{c}}=\vec{\text{c}}\times\vec{\text{a}}\Rightarrow\Bigg[\vec{\text{a}}\times\vec{\text{b}}\Bigg]=\Bigg[\vec{\text{b}}\times\vec{\text{c}}\Bigg]=\Bigg[\vec{\text{c}}\times\vec{\text{a}}\Bigg]$
$\therefore\text{ab sin C = bc sin A = ca sin B }\text{ Or }\frac{\sin\text{C}}{\text{c}}=\frac{\sin\text{A}}{\text{a}}=\frac{\sin\text{B}}{\text{b}}$
$\therefore\frac{\text{a}}{\text{sin A}}=\frac{\text{b}}{\text{sin B}}=\frac{\text{c}}{\text{sin C}}.$ View full question & answer→Question 53 Marks
Find the projection of $\overrightarrow{b} + \overrightarrow{c} $ on $\overrightarrow{a}$ where $\overrightarrow{a} = 2\hat{i} - 2\hat{j} + \hat{k}, \overrightarrow{b} = \hat{i} + 2\hat{j} - 2\hat{k} $ and $\overrightarrow{c} = 2\hat{i} - \hat{j} + 4\hat{k}.$
Answer$\overrightarrow{b} + \overrightarrow{c} = 3\hat{i} + \hat{j}+ 2\hat{k} ;\overrightarrow{a} =2\hat{i} - 2\hat{j}+\hat{k} $$\therefore \text{Projection of} (\overrightarrow{b}+\overrightarrow{c}) \text{on} \overrightarrow{a} = \frac{\bigg(3\hat{i}+\hat{j}+2\hat{k}\bigg).{\bigg(2\hat{i} -2\hat{j}+\hat{k}\bigg)}}{\bigg|2\hat{i}-2\hat{j}+\hat{k}\bigg|} $
$\frac{6 - 2 + 2}{3} = 2$
View full question & answer→Question 63 Marks
Find the value of $\lambda$ $\overrightarrow{a} ,\overrightarrow{b}$ and $\overrightarrow{c}$ coplanar, where $\overrightarrow{a} = 2\hat{i} -\hat{j} + \hat{k}, \overrightarrow{b} = \hat{i} + 2\hat{j} - 3\hat{k} $ and $\overrightarrow{c} = 3\hat{i}- \lambda \hat{j} + 5\hat{k}.$
Answer$\overrightarrow{a} = 2\hat{i} -\hat{j} + \hat{k}, \overrightarrow{b} = \hat{i} + 2\hat{j} - 3\hat{k}, \overrightarrow {c} = 3\hat{i}- \lambda \hat{j} + 5\hat{k}.$ are coplanar if $\bigg[\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c} \bigg] = 0$
$ \begin{vmatrix} 2 & -1 & 1 \\ 1 & 2 & -3 \\ 3 & -\lambda & 5 \end{vmatrix} = 0 $
$\Rightarrow 2 (10 - 3 \lambda) + 1 (5 + 9) + ( -\lambda - 6) =0$
$\Rightarrow 7 \lambda = 28$
OR
$\lambda = 4$
View full question & answer→Question 73 Marks
Using vectors prove that the line segment joining the mid-points of non-parallel sides of a trapezium is parallel to the base and is equal to half the sum of the parallel sides.
AnswerLet $\overrightarrow{a}, \overrightarrow{b} \text{and}\overrightarrow{c}$ be the position vectors of A, B and C respectively w.r.t. O. Let D and E be the mid-points of parallel sides OC and AB respectively.
$\therefore \text{Position vector of}$
$\text{D is} \frac{O +\overrightarrow{C}}{2} = \frac{\overrightarrow{c}}{2}$
$CB\parallel OA \Rightarrow \overrightarrow{a}\parallel \overrightarrow{b} - \overrightarrow{c}$
$\Rightarrow \text{E is parallel to base and its length is half the sum of lengths of the parallel sides.}$ View full question & answer→Question 83 Marks
$\overrightarrow{a} = \hat{i} + 2\hat{j} - 3\hat{k}, \overrightarrow{b} = 3\hat{i} - \hat{j} + 2\hat{k}, \text{show that}\bigg(\overrightarrow{a} +\overrightarrow{b}\bigg) \text{and} \bigg(\overrightarrow{a} -\overrightarrow{b}\bigg)$ are perpendicular to each other.
Answer$(\overrightarrow{a} +\overrightarrow{b}) = 4\hat{i} + \hat{j} - \hat{k} ,\ (\overrightarrow{a} -\overrightarrow{b}) = - 2\hat{i} + 3\hat{j} -5\hat{k}$$\text{For} (\overrightarrow{a} +\overrightarrow{b})\ and\ (\overrightarrow{a} -\overrightarrow{b}) \text{ to be perpendicular } (\overrightarrow{a}+\overrightarrow{b}).(\overrightarrow{a} -\overrightarrow{b})$
$(\overrightarrow{a} +\overrightarrow{b}).(\overrightarrow{a}-\overrightarrow{b}) = 4(-2) + 1.3 + (-1)(-5)$
$= - 8 + 3 + 5 = 0$
$\therefore (\overrightarrow{a}+\overrightarrow{b}) \perp (\overrightarrow{a}-\overrightarrow{b})$
View full question & answer→Question 93 Marks
Prove that the sum of all vectors drawn from the centre of a regular octagon to its vertices is the zero vector.
AnswerGiven: A regular octagon of eight sides with center O.
To show: $\overrightarrow{\text{OA}}+\overrightarrow{\text{OB}}+\overrightarrow{\text{OC}}+\overrightarrow{\text{OD}}+\overrightarrow{\text{OE}}+\overrightarrow{\text{OF}}+\overrightarrow{\text{OG}}+\overrightarrow{\text{OH}}=\vec0$.
Proof: We know center of the regular octagon bisects all the diagonals passing through it.
$\overrightarrow{\text{OA}}=-\overrightarrow{\text{OE}},\ \overrightarrow{\text{OB}}=-\overrightarrow{\text{OF}},\ \overrightarrow{\text{OD}}=-\overrightarrow{\text{OH}}$ and $\overrightarrow{\text{OC}}=-\overrightarrow{\text{OG}}.$
$\Rightarrow\overrightarrow{\text{OA}}+\overrightarrow{\text{OE}}=\vec0,\ \overrightarrow{\text{OB}}+\overrightarrow{\text{OF}}=\vec0,\ \overrightarrow{\text{OD}}+\overrightarrow{\text{OH}}=\vec0$ and $\overrightarrow{\text{OC}}+\overrightarrow{\text{OG}}=\vec0.\ \dots(\text{i})$
Now,
$\overrightarrow{\text{OA}}+\overrightarrow{\text{OB}}+\overrightarrow{\text{OC}}+\overrightarrow{\text{OD}}+\overrightarrow{\text{OE}}+\overrightarrow{\text{OF}}+\overrightarrow{\text{OG}}+\overrightarrow{\text{OH}}$
$=\Big(\overrightarrow{\text{OA}}+\overrightarrow{\text{OE}}\Big)+\Big(\overrightarrow{\text{OB}}+\overrightarrow{\text{OF}}\Big)+\Big(\overrightarrow{\text{OC}}+\overrightarrow{\text{OG}}\Big)+\Big(\overrightarrow{\text{OD}}+\overrightarrow{\text{OH}}\Big)$
$=\vec0+\vec0+\vec0+\vec0$
$=\vec0$
Hence proved.
View full question & answer→Question 103 Marks
Find the vector from the origin O to the centroid of the triangle whose vertices are (1, -1, 2), (2, 1, 3) and (-1, 2, -1).
AnswerGiven the vertices of the triangle (1, -1, 2), (2, 1, 3) and (-1, 2, -1). Then, Position vectors are: $\vec{\text{a}}=\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}$ $\vec{\text{b}}=2\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k}}$ $\vec{\text{c}}=-\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}$ The centroid of a triangle is given by $\frac{\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}}{3}$ So, $\frac{\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}}{3}=\frac{\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}+2\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k}}-\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}}3{}$ $=\frac{2\hat{\text{i}}+2\hat{\text{j}}+4\hat{\text{k}}}3=\frac{2}3\hat{\text{i}}+\frac{2}3\hat{\text{j}}+\frac{4}3\hat{\text{k}}$
View full question & answer→Question 113 Marks
Using vectors, find the value of k such that the points A(k, -10, 3), (1, -1, 3) and (3, 5, 3) are collinear.
AnswerLet the points are A(k, -10, 3), B(1, -1, 3) and C(3, 5, 4)
$\therefore\overrightarrow{\text{AB}}=\overrightarrow{\text{OB}}-\overrightarrow{\text{OA}}$
$=(\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}})-(\text{k}\hat{\text{i}}-10\hat{\text{j}}+3\hat{\text{k}})$
$=(1+\text{k})\hat{\text{i}}+9\hat{\text{j}}$
and $\overrightarrow{\text{BC}}=\overrightarrow{\text{OC}}-\overrightarrow{\text{OB}}$
$=(3\hat{\text{i}}-5\hat{\text{j}}+3\hat{\text{k}})-(\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}})$
$=2\hat{\text{i}}+6\hat{\text{j}}$
If point A, B, C are collinear then there exists some scalar $'\lambda'$ such that
$\Rightarrow\overrightarrow{\text{AB}}=\lambda\overrightarrow{\text{BC}}$
$(1-\text{k})\hat{\text{i}}+9\hat{\text{j}}=\lambda(2\hat{\text{i}}+6\hat{\text{j}})$
Comparing coefficients we get $1-\lambda=2\lambda$ and $9=6\lambda$
$3-3\text{k}=9$
$\text{k}=-2$
View full question & answer→Question 123 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 133 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 143 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 153 Marks
If the position vectors of the points A(3, 4), B(5, -6) and C(4, -1) are $\vec{\text{a}},\ \vec{\text{b}},\ \vec{\text{c}}$ respectively, compute $\vec{\text{a}}+2\vec{\text{b}}-3\vec{\text{c}}$.
AnswerHere, A(3, 4), B(5, -6), C(4, -1) $\vec{\text{a}}=3\hat{\text{i}}+4\hat{\text{j}}$ $\vec{\text{b}}=5\hat{\text{i}}-6\hat{\text{j}}$ $\vec{\text{c}}=4\hat{\text{i}}-\hat{\text{j}}$ Now,$\vec{\text{a}}+2\vec{\text{b}}-3\vec{\text{c}}=\big(3\hat{\text{i}}+4\hat{\text{j}}\big)+2\big(5\hat{\text{i}}-6\hat{\text{j}}\big)-3\big(4\hat{\text{i}}-\hat{\text{j}}\big)$
$=3\hat{\text{i}}+4\hat{\text{j}}+10\hat{\text{i}}-12\hat{\text{j}}-12\hat{\text{i}}+3\hat{\text{j}}$ $=\hat{\text{i}}-5\hat{\text{j}}$ $\therefore\ \vec{\text{a}}+2\vec{\text{b}}-3\vec{\text{c}}=\hat{\text{i}}-5\hat{\text{j}}$
View full question & answer→Question 163 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 173 Marks
Find a unit vector perpendicular to both the vectors $4\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}$ and $-2\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}.}$
AnswerA vector perpendicular to both $\vec{\text{a}}$ and $\vec{\text{b}}=\vec{\text{a}}\times\vec{\text{b}}.$
$\vec{\text{a}}\times\vec{\text{b}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\\text{a}_1&\text{b}_1&\text{c}_1\\\text{a}_2&\text{b}_2&\text{c}_2 \end{vmatrix}$
$\vec{\text{c}}(\text{say})=\begin{bmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\4&-1&3\\-2&1&-2 \end{bmatrix}$
$\vec{\text{c}}=\hat{\text{i}}(2-3)-\hat{\text{j}}(-8+6)+\hat{\text{k}}(4-2)$
$\vec{\text{c}}=-\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}$
$\vec{\text{c}}$ is a vector perpendicular to both $\vec{\text{a}}$ and $\vec{\text{b}}.$
$\hat{\text{c}}=\frac{\vec{\text{c}}}{|\vec{\text{c}}|}$
$=\frac{-\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}}{\sqrt{(-1)^2+(2)^2+(2)^2}}$
$=\frac{-\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}}{\sqrt{1+4+4}}$
$=\frac{1}{3}\big(-\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}\big)$
So, unit vector perpendicular to both $\vec{\text{a}}$ and $\vec{\text{b}}=\frac{1}{3}\big(-\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}\big).$
View full question & answer→Question 183 Marks
ABCDE is a pentagon, prove that,$\overrightarrow{\text{AB}}+\overrightarrow{\text{AE}}+\overrightarrow{\text{BC}}+\overrightarrow{\text{DC}}+\overrightarrow{\text{ED}}+\overrightarrow{\text{AC}}=3\ \overrightarrow{\text{AC}}$
AnswerIt is given that ABCDE is a pentagon, So
$\overrightarrow{\text{AB}}+\overrightarrow{\text{AE}}+\overrightarrow{\text{BC}}+\overrightarrow{\text{DC}}+\overrightarrow{\text{ED}}+\overrightarrow{\text{AC}}$
$=\Big(\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}\Big)+\overrightarrow{\text{AE}}+\overrightarrow{\text{DC}}+\overrightarrow{\text{ED}}+\overrightarrow{\text{AC}}$
$=\overrightarrow{\text{AC}}+\overrightarrow{\text{DC}}+\Big(\overrightarrow{\text{AE}}+\overrightarrow{\text{ED}}\Big)+\overrightarrow{\text{AC}}$ $\Big[$Using triangle law in $\triangle\text{ABC},\ \overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}=\overrightarrow{\text{AC}}\Big]$
$=\overrightarrow{\text{AC}}+\overrightarrow{\text{DC}}+\Big(\overrightarrow{\text{AD}}\Big)+\overrightarrow{\text{AC}}$
$=\overrightarrow{\text{AC}}+\overrightarrow{\text{DC}}-\overrightarrow{\text{DA}}+\overrightarrow{\text{AC}}$
$=\overrightarrow{\text{AC}}+\overrightarrow{\text{DC}}+\overrightarrow{\text{AD}}+\overrightarrow{\text{AC}}$
$=\overrightarrow{\text{AC}}+\overrightarrow{\text{AD}}+\overrightarrow{\text{DC}}+\overrightarrow{\text{AC}}$
$=\overrightarrow{\text{AC}}+\overrightarrow{\text{AC}}+\overrightarrow{\text{AC}}$
$=3\ \overrightarrow{\text{AC}}$
So,
$\overrightarrow{\text{AB}}+\overrightarrow{\text{AE}}+\overrightarrow{\text{BC}}+\overrightarrow{\text{DC}}+\overrightarrow{\text{ED}}+\overrightarrow{\text{AC}}=3\ \overrightarrow{\text{AC}}$
View full question & answer→Question 193 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 203 Marks
Write the position vector of the point which divdes the join of the points with position vectors $3\vec{\text{a}} - 2\vec{\text{b}} $ and $2\vec{\text{a}} + 3\vec{\text{b}} $ in the ratio 2 : 1.
AnswerSuppose R be the point which divdes the line joining the points with position vectors $3\vec{\text{a}} - 2\vec{\text{b}} $ and $2\vec{\text{a}} + 3\vec{\text{b}} $ in the ratio 2 : 1
And $\overrightarrow{\text{OA}} = 3\vec{\text{a}} - 2\vec{\text{b}} $ and $\overrightarrow{\text{OB}} = 2\vec{\text{a}} + 3\vec{\text{b}} $
Here, m : n = 2 : 1
Therefore, position vector $\overrightarrow{\text{OR}}$ is as follows:
$\overrightarrow{\text{OR}} = \frac{\text{m} \overrightarrow{\text{OB}} + \text{n} \overrightarrow{\text{OA}}}{\text{m + n}}$
$ = \frac{2\big(2\vec{\text{a}} + 3 \vec{\text{b}}\big) + 1 \big(3\vec{\text{a}} + 2 \vec{\text{b}}\big)}{2 + 1}$
$= \frac{7\vec{\text{a}} + 4\vec{\text{b}}}{3}$
$= \frac{7}{3} \vec{\text{a}} + \frac{4}{3} \vec{\text{b}}$
View full question & answer→Question 213 Marks
If $\overrightarrow{\text{PQ}}=3\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}$ and the coordinates of P are (1, -1, 2), find the coordinates of Q.
AnswerHere, $\overrightarrow{\text{PQ}}=3\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}$
Position vector of $\text{P}=\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}$
$\overrightarrow{\text{PQ}}=$ Position vector of Q - Position vector of P
$3\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}=$ Position vector of Q $-\big(\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}\big)$
Position vector of $\text{Q}=\big(\hat{3\text{i}}+2\hat{\text{j}}-\hat{\text{k}}\big)+\big(\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}\big)$
$=4\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$
Coordinates of Q = (4, 1, 1)
View full question & answer→Question 223 Marks
A vector $\vec{\text{r}}$ is inclined at equal angles to the three axes. If the magnitude of $\vec{\text{r}}$ is $2\sqrt3$, find $\vec{\text{r}}$.
AnswerLet l, m, n be the direction cosines of $\vec{\text{r}}$.
Now, $\vec{\text{r}}$ is inclined at equal angles to the three axes.
$\therefore\ \text{l}=\text{m}=\text{n}$ $[\alpha=\beta=\gamma\Rightarrow\cos\alpha=\cos\beta=\cos\gamma]$
So, $\text{l}^2+\text{m}^2+\text{n}^2=1$
$\Rightarrow\ 3\text{l}^2=1$
$\Rightarrow\ \text{l}=\pm\frac{1}{\sqrt3}$
$\Rightarrow\ \text{l = m = n}=\pm\frac{1}{\sqrt3}$
We know that,
$\vec{\text{r}}=|\vec{\text{r}}|(\text{l}\hat{\text{i}}+\text{m}\hat{j}+\text{n}\hat{\text{k}})$
$\Rightarrow\ \vec{\text{r}}=2\sqrt3\Big(\pm\frac{1}{\sqrt3}\hat{\text{i}}\pm\frac{1}{\sqrt3}\hat{\text{j}}\pm\frac{1}{\sqrt3}\hat{\text{k}}\Big)$
$\Rightarrow\ \vec{\text{r}}=2\big(\pm\hat{\text{i}}\pm\hat{\text{j}}\pm\hat{\text{k}}\big)$
View full question & answer→Question 233 Marks
Write the number of vectors of unit length perpendicular to both the vectors $\vec{\text{a}}=2\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}$and $\vec{\text{b}}=\hat{\text{j}}+\hat{\text{k}}.$
AnswerUnit vectors perpendicular to $\vec{\text{a}}$ and $\vec{\text{b}}$ are $\pm\Bigg(\frac{\vec{\text{a}}\times\vec{\text{b}}}{\big|\vec{\text{a}}\times\vec{\text{b}}\big|}\Bigg).$
$\vec{\text{a}}\times\vec{\text{b}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\2&1&2\\0&1&1\end{vmatrix}=-\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}}$
$\therefore$ Unit vectors perpendicular to $\vec{\text{a}}$ and $\vec{\text{b}}$ are $\pm\frac{-\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}}}{\sqrt{(-1)^2+(-2)^2+(2)^2}}=\pm\Big(-\frac{1}{3}\hat{\text{i}}-\frac{2}{3}\hat{\text{j}}+\frac{2}{3}\hat{\text{k}}\Big)$
Thus, there are two unit vectors perpendicular to the given vectors.
View full question & answer→Question 243 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 253 Marks
ABCD is a parallelogram. If the coordinates of A, B, C are (-2, -1), (3, 0) and (1, -2) respectively, find the coordinates of D.
AnswerLet the coordinates of D is (x, y).
Since, ABCD is a parallelogram.
$\therefore$ AB = DC
We have,
$\overrightarrow{\text{AB}}=\overrightarrow{\text{DC}}$
$\Rightarrow3\hat{\text{i}}-\big(-2\hat{\text{i}}-\hat{\text{j}}\big)=\big(\hat{\text{i}}-2\hat{\text{j}}\big)-\big(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}\big)$
$\Rightarrow5\hat{\text{i}}+\hat{\text{j}}=\hat{\text{i}}(1-\text{x})+\hat{\text{j}}(-2-\text{y})$
$\Rightarrow1-\text{x}=5\text{ and }1=-2-\text{y}$
$\Rightarrow\text{x}=-4\text{ and }\text{y}=-3$
Hence, the coordinates of D is (-4, -3)
View full question & answer→Question 263 Marks
If the position vector $\vec{\text{a}}$ of a point (12, n) is such that $\big|\vec{\text{a}}\big|=13$, find the value (s) of n.
AnswerGiven a position vector $\vec{\text{a}}$ of a point (12, n) such that,
$\vec{\text{a}}=12\hat{\text{i}}+\text{n}\hat{\text{j}}$
Then,
$\big|\vec{\text{a}}\big|=\sqrt{12^2+\text{n}^2}$
Also, $\big|\vec{\text{a}}\big|=13$ (given)
Thus, we get,
$\sqrt{12^2+\text{n}^2}=13$
$\Rightarrow12^2+\text{n}^2=169$
$\Rightarrow\text{n}^2=169-144$
$\Rightarrow\text{n}^2=25$
$\Rightarrow\text{n}=\pm5$
View full question & answer→Question 273 Marks
If $|\vec{\text{a}}|=\sqrt{26,}\big|\vec{\text{b}}\big|=7$ and $\big|\vec{\text{a}}\times\vec{\text{b}}\big|=35,$ find $\vec{\text{a}}.\vec{\text{b}}.$
Answer$\vec{\text{a}}\times\vec{\text{b}}=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\sin\theta.\hat{\text{n}}$
$\big|\vec{\text{a}}\times\vec{\text{b}}\big|=|\vec{\text{a}}|\big|\vec{\text{b}}\big||\sin\theta||\hat{\text{n}}|$
$35=\sqrt{26.7}|\sin\theta|.1$
$\sin\theta=\frac{35}{\sqrt{26.5}}$
$\sin\theta=\frac{5}{\sqrt{26}}$
$\cos^2\theta=1-\sin^2\theta$
$=1-\Big(\frac{5}{\sqrt{26}}\Big)^2$
$=\frac{1}{1}-\frac{25}{26}$
$=\frac{26-25}{26}$
$=\frac{1}{26}$
$\cos\theta=\frac{1}{\sqrt{26}}$
$\vec{\text{a}}.\vec{\text{b}}=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta$
$=\sqrt{26}.7.\frac{1}{\sqrt{26}}$
$\vec{\text{a}}.\vec{\text{b}}=7$
View full question & answer→Question 283 Marks
If $\vec{\text{a}}\times\vec{\text{b}}=\vec{\text{b}}\times\vec{\text{c}}\neq0,$ then show that $\vec{\text{a}}+\vec{\text{c}}=\text{m}\vec{\text{b}},$ where m is any scalar.
Answer$\vec{\text{a}}\times\vec{\text{b}}=\vec{\text{b}}\times\vec{\text{c}}$$\Rightarrow\vec{\text{a}}\times\vec{\text{b}}=-\vec{\text{c}}\times\vec{\text{b}}$
$\Rightarrow\vec{\text{a}}\times\vec{\text{b}}+\vec{\text{c}}\times\vec{\text{b}}=0$
$\Rightarrow\big(\vec{\text{a}}+\vec{\text{c}}\big)\times\vec{\text{b}}=0$ (using right distributive property)
Thus, $\vec{\text{a}}+\vec{\text{c}}$ is parallel to $\vec{\text{b}}.$
$\Leftrightarrow\vec{\text{a}}+\vec{\text{c}}=\text{m}\vec{\text{b}},$ for som scalar m.
View full question & answer→Question 293 Marks
Define $\vec{\text{a}}\times\vec{\text{b}}$ and prove that $\big|\vec{\text{a}}\times\vec{\text{b}}\big|=\big(\vec{\text{a}}.\vec{\text{b}}\big)\tan\theta,$ where $\theta$ is the angle between $\vec{\text{a}}$ and $\vec{\text{b}}.$
AnswerIf $\vec{\text{a}}$ and $\vec{\text{b}}$ are two non-parallel vectors, then the vectors product denoted by $\vec{\text{a}}\times\vec{\text{b}}$ is defined as $\vec{\text{a}}\times\vec{\text{b}}=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\sin\theta\hat{\text{ n}}.$
Here, $\theta$ is the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$ and $\hat{\text{n}}$ is the unit vector perpendicular to the plane of $\vec{\text{a}}$ and $\vec{\text{b}}$ such that $\vec{\text{a}},\vec{\text{b}}$ and $\hat{\text{n}}$ form a right
$\text{LHS}=\big|\vec{\text{a}}\times\vec{\text{b}}\big|$
$=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\sin\theta$
$=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\sin\theta\times\frac{\cos\theta}{\cos\theta}$
$=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta\frac{\sin\theta}{\cos\theta}$
$=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta\tan\theta$
$=\big(\vec{\text{a}}.\vec{\text{b}}\big)\tan\theta$
$=\text{RHS}$
Hence proved
View full question & answer→Question 303 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 313 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 323 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 333 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 343 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 353 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 363 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 373 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 383 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 393 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 403 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 413 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 423 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 433 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 443 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 453 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→Question 463 Marks
If $\vec{\text{a}},\vec{\text{b}}$ are two vectors, then write the truth value of the following statement:$\big|\vec{\text{a}}\big|=\big|\vec{\text{b}}\big|\Rightarrow\vec{\text{a}}=\pm\vec{\text{b}}$
AnswerFalse
$\big|\vec{\text{a}}\big|=\big|\vec{\text{b}}\big|\Rightarrow\vec{\text{a}}=\pm\vec{\text{b}}$
Consider an example,
$\vec{\text{a}}=\text{i}+\sqrt3\text{j}$ and $\vec{\text{b}}=\sqrt2\text{i}+\sqrt2\text{j}$
$\big|\vec{\text{a}}\big|=\sqrt{1^2+\big(\sqrt3\big)^2}=2$ and $\big|\vec{\text{b}}\big|=\sqrt{\big(\sqrt2\big)^2+\big(\sqrt2\big)^2}=2$
Thus, $\big|\vec{\text{a}}\big|=\big|\vec{\text{b}}\big|$ but $\vec{\text{a}}\neq\pm\vec{\text{b}}$
View full question & answer→Question 473 Marks
Express $\overrightarrow{\text{AB}}$ in terms of unit vectors $\hat{\text{i}}\text{ and }\hat{\text{j}}$, when the point is:A(4, -1), B(1, 3)
Find $\Big|\overrightarrow{\text{AB}}\Big|$
AnswerHere, A = (4, -1) B = (1, 3) Position vector of $\text{A}=4\hat{\text{i}}-\hat{\text{j}}$ Position vector of $\text{B}=\hat{\text{i}}+3\hat{\text{j}}$ $\overrightarrow{\text{AB}}$ = Position vector of B - Position vector of A$=\big(\hat{\text{i}}+3\hat{\text{j}}\big)-\big(4\hat{\text{i}}-\hat{\text{j}}\big)$
$=\hat{\text{i}}+3\hat{\text{j}}-4\hat{\text{i}}+\hat{\text{j}}$
$\overrightarrow{\text{AB}}=-3\hat{\text{i}}+4\hat{\text{j}}$
$\Big|\overrightarrow{\text{AB}}\Big|=\sqrt{(-3)^2+(4)^2}$
$=\sqrt{9+16}$
$=\sqrt{25}$
$\Big|\overrightarrow{\text{AB}}\Big|=5$
$\overrightarrow{\text{AB}}=-3\hat{\text{i}}+4\hat{\text{j}}$
View full question & answer→Question 483 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 493 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 503 Marks
Find the magnitude of $\vec{\text{a}}=\big(3\hat{\text{k}}+4\hat{\text{j}}\big)\times(\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}).$
Answer$\vec{\text{a}}=\big(0\hat{\text{i}}+4\hat{\text{j}}+3\hat{\text{k}}\big)\times\big(\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}\big)$
$=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\0&4&3\\1&1&-1 \end{vmatrix}$
$=\hat{\text{i}}(-4-3)-\hat{\text{j}}(0-3)+\hat{\text{k}}(0-4)$
$=-7\hat{\text{i}}+3\hat{\text{j}}-4\hat{\text{k}}$
$\Rightarrow|\vec{\text{a}}|=\sqrt{(-7)^2+3^2+(-4)^2}$
$=\sqrt{74}$
View full question & answer→