Find the equation of the line in vector and in Cartesian form that passes through the point with position vector
Question: Find the equation of the line in vector and in Cartesian form that passes through the point with position vector $2 \hat{i}-\hat{j}+4 \hat{k}$ and is in the direction $\hat{i}+2 \hat{j}-\hat{k}$. Solution: It is given that the line passes through the point with position vector $\vec{a}=2 \hat{i}-\hat{j}+4 \hat{k}$ ...(1) $\vec{b}=\hat{i}+2 \hat{j}-\hat{k}$ $\ldots(2)$ It is known that a line through a point with position vector $\vec{a}$ and parallel to $\vec{b}$ is given by the equati...
Read More →Duration of sunshine(in hours) in Amritsar for first 10 days of August 1997 as reported
Question: Duration of sunshine(in hours) in Amritsar for first 10 days of August 1997 as reported by the Meterological Department are given as follows : 9.6 , 5.2 , 3.5 , 1.5 , 1.6 , 2.4 , 2.6 , 8.4 , 10.3 , 10.9 1. Find the mean $\overline{\mathrm{X}}$ 2. Verify that $\sum_{i=1}^{10}\left(x^{i}-\bar{X}\right)=0$ Solution: Duration of sunshine (in hours ) for 10 days are = 9.6 , 5.2 , 3.5 , 1.5 , 1.6 , 2.4 , 2.6 , 8.4 , 10.3 , 10.9 (i) Mean $X=\frac{\text { Sum of numbers }}{\text { Total number...
Read More →If M is the mean of x1, x2,x3,x4,x5 and x6, Prove that
Question: If $M$ is the mean of $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ and $x_{6}$, Prove that $\left(x_{1}-M\right)+\left(x_{2}-M\right)+\left(x_{3}-M\right)+\left(x_{4}-M\right)+\left(x_{5}-M\right)+\left(x_{6}-M\right)=0$. Solution: Let $M$ be the mean of $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ and $x_{6}$ Then, $\mathrm{M}=\frac{\mathrm{x}_{1}+\mathrm{x}_{2}+\mathrm{x}_{3}+\mathrm{x}_{4}+\mathrm{x}_{5}+\mathrm{x}_{6}}{6}$ $=x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}=6 M$ To Prove :- $\left(x_{1}-M\right)+...
Read More →Find the equation of the line which passes through the point (1, 2, 3) and is parallel to the vector.
Question: Find the equation of the line which passes through the point (1, 2, 3) and is parallel to the vector. Solution: It is given that the line passes through the point $A(1,2,3)$. Therefore, the position vector through $A$ is $\vec{a}=\hat{i}+2 \hat{j}+3 \hat{k}$ $\vec{b}=3 \hat{i}+2 \hat{j}-2 \hat{k}$\ It is known that the line which passes through point $\mathrm{A}$ and parallel to $\vec{b}$ is given by $\vec{r}=\vec{a}+\lambda \vec{b}$, where $\lambda$ is a constant. $\Rightarrow \vec{r}...
Read More →Prove that sin
Question: Prove that $\sin 3 x+\sin 2 x-\sin x=4 \sin x \cos \frac{x}{2} \cos \frac{3 x}{2}$ Solution: LHS $=\sin 3 x+\sin 2 x-\sin x$ $=\sin 3 x+2 \sin \left(\frac{2 x-x}{2}\right) \cos \left(\frac{2 x+x}{2}\right) \quad\left[\because \sin A-\sin B=2 \sin \left(\frac{A-B}{2}\right) \cos \left(\frac{A+B}{2}\right)\right]$ $=\sin 3 x+2 \sin \left(\frac{x}{2}\right) \cos \left(\frac{3 x}{2}\right)$ $=2 \sin \left(\frac{3 x}{2}\right) \cos \left(\frac{3 x}{2}\right)+2 \sin \left(\frac{3 x}{2}\right...
Read More →The numbers of children in 10 families of a locality are
Question: The numbers of children in 10 families of a locality are 2, 4, 3, 4, 2, 3, 5, 1, 1, 5. Find the number of children per family. Solution: The numbers of children in 10 families are : 2, 4, 3, 4, 2, 3, 5, 1, 1, 5 $\therefore$ Mean $=\frac{\text { Total no. children }}{\text { Total families }}$ $=\frac{2+4+3+4+2+3+5+1+1+5}{10}=3$...
Read More →If sec A=54, verify that 3 sin A−4 sin3 A4 cos3 A−3 cos A=3 tan A−tan3 A1−3 tan2 A.
Question: If sec $A=\frac{5}{4}$, verify that $\frac{3 \sin A-4 \sin ^{3} A}{4 \cos ^{3} A-3 \cos A}=\frac{3 \tan A-\tan ^{3} A}{1-3 \tan ^{2} A}$. Solution: Given: $\sec A=\frac{5}{4}$....(1) To verify: $\frac{3 \sin A-4 \sin ^{3} A}{4 \cos ^{3} A-3 \cos A}=\frac{3 \tan A-\tan ^{3} A}{1-3 \tan ^{2} A}$.... (2) Now we know that $\sec A=\frac{1}{\cos A}$ Therefore $\cos A=\frac{1}{\sec A}$ Now, by substituting the value of $\sec A$ from equation (1) We get, $\cos A=\frac{1}{\frac{5}{4}}$ $=\frac{...
Read More →Let AB be the line through the points, (4, 7, 8) and (2, 3, 4), and CD be the line through the points,
Question: Show that the line through the points (4, 7, 8) (2, 3, 4) is parallel to the line through the points (1, 2, 1), (1, 2, 5). Solution: Let AB be the line through the points, (4, 7, 8) and (2, 3, 4), and CD be the line through the points, (1, 2, 1) and (1, 2, 5). The directions ratios, $a_{1}, b_{1}, c_{1}$, of $A B$ are $(2-4),(3-7)$, and $(4-8)$ i.e., $-2,-4$, and $-4$ The direction ratios, $a_{2}, b_{2}, c_{2}$, of CD are $(1-(-1)),(2-(-2))$, and $(5-1)$ i.e., 2,4, and 4 . AB will be p...
Read More →Prove that:
Question: Prove that: $\cos 4 x-\cos 4 \alpha=8(\cos x-\cos \alpha)(\cos x+\cos \alpha)(\cos x-\sin \alpha)(\cos x+\sin \alpha)$ Solution: RHS $=8(\cos x-\cos \alpha)(\cos x+\cos \alpha)(\cos x-\sin \alpha)(\cos x+\sin \alpha)$ $=8\left(\cos ^{2} x-\cos ^{2} \alpha\right)\left(\cos ^{2} x-\sin ^{2} \alpha\right)$ $=8\left(\cos ^{4} x-\cos ^{2} x \times \sin ^{2} \alpha-\cos ^{2} \alpha \times \cos ^{2} x+\cos ^{2} \alpha \times \sin ^{2} \alpha\right)$ $=8\left\{\cos ^{4} x-\cos ^{2} x\left(\sin...
Read More →The percentage marks obtained by students of a class in mathematics are as follows:
Question: The percentage marks obtained by students of a class in mathematics are as follows: 64, 36 , 47, 23, 0, 19, 81, 93, 72, 35, 3, 1 .Find their mean. Solution: The percentage marks obtained by students are 64, 36 , 47, 23, 0, 19, 81, 93, 72, 35, 3, 1 $\therefore$ Mean marks $=\frac{\text { Sum of marks }}{\text { Total numbers of marks }}$ $=\frac{64+36+47+23+0+19+81+93+72+35+3+1}{5}=39.5$ Mean Marks = 39.5...
Read More →Following are the weights of 10 new born babies in a hospital on a particular day:
Question: Following are the weights of 10 new born babies in a hospital on a particular day: 3.4 , 3 .6 , 4.2 , 4.5 , 3.9 , 4.1 , 3.8 , 4.5 , 4.4 , 3.6 (in kg). Find the mean. Solution: The weights (in kg) of 10 new born babies are : 3.4 , 3 .6 , 4.2 , 4.5 , 3.9 , 4.1 , 3.8 , 4.5 , 4.4 , 3.6 $\therefore$ Mean Weight $=\frac{\text { Sum of weights }}{\text { Total no. of babies }}$ $=\frac{3.4+3.6+4.2+4.5+3.9+4.1+3.8+4.5+4.4+3.6}{10}$ = 4 kg...
Read More →Show that the line through the points (1, −1, 2) (3, 4, −2) is perpendicular to the line through the points (0, 3, 2) and (3, 5, 6).
Question: Show that the line through the points (1, 1, 2) (3, 4, 2) is perpendicular to the line through the points (0, 3, 2) and (3, 5, 6). Solution: Let AB be the line joining the points, (1, 1, 2) and (3, 4, 2), and CD be the line joining the points, (0, 3, 2) and (3, 5, 6). The direction ratios, $a_{1}, b_{1}, c_{1}$, of AB are $(3-1),(4-(-1))$, and $(-2-2)$ i.e., 2,5, and $-4$. The direction ratios, $a_{2}, b_{2}, c_{2}$, of CD are $(3-0),(5-3)$, and $(6-2)$ i.e., 3,2 , and 4 . $A B$ and $C...
Read More →Find the mean of first five multiples of 3.
Question: Find the mean of first five multiples of 3. Solution: First five multiples of 3 are 3, 6, 9, 12, 15. $\therefore$ Mean $=\frac{\text { Sum of numbers }}{\text { Total numbers }}$ $=\frac{3+6+9+12+15}{5}=\frac{45}{5}$ = 9 Mean = 9...
Read More →Prove that:
Question: Prove that: $\cot ^{2} x-\tan ^{2} x=4 \cot 2 x \operatorname{cosec} 2 x$ Solution: $\mathrm{LHS}=\cot ^{2} x-\tan ^{2} x$ $=\frac{\cos ^{2} x}{\sin ^{2} x}-\frac{\sin ^{2} x}{\cos ^{2} x}$ $=\frac{\left(\cos ^{2} x\right)^{2}-\left(\sin ^{2} x\right)^{2}}{\sin ^{2} x \cos ^{2} x}$ $=\frac{\left(\cos ^{2} x+\sin ^{2} x\right)\left(\cos ^{2} x-\sin ^{2} x\right)}{\sin ^{2} x \cos ^{2} x}$ $=\frac{1 \times(\cos 2 x)}{\sin ^{2} x \cos ^{2} x} \quad\left(\because \cos ^{2} x-\sin ^{2} x=\c...
Read More →Show that the three lines with direction cosines
Question: Show that the three lines with direction cosines $\frac{12}{13}, \frac{-3}{13}, \frac{-4}{13} ; \frac{4}{13}, \frac{12}{13}, \frac{3}{13} ; \frac{3}{13}, \frac{-4}{13}, \frac{12}{13}$ are mutually perpendicular. Solution: Two lines with direction cosines,l1,m1,n1andl2,m2,n2, are perpendicular to each other, ifl1l2+m1m2+n1n2= 0 (i) For the lines with direction cosines, $\frac{12}{13}, \frac{-3}{13}, \frac{-4}{13}$ and $\frac{4}{13}, \frac{12}{13}, \frac{3}{13}$, we obtain $l_{1} l_{2}+m...
Read More →Find the mean of x, x + 2, x + 4, x + 6, x + 8.
Question: Find the mean of x, x + 2, x + 4, x + 6, x + 8. Solution: Numbers are x, x + 2, x + 4, x + 6, x + 8. $\therefore$ Mean $=\frac{\text { Sum of numbers }}{\text { Total numbers }}$ $=\frac{x+x+2+x+4+x+6+x+8}{5}$ $=\frac{5 x+20}{5}$ $=5\left(\frac{x+4}{5}\right)$ = x + 4...
Read More →Prove that:
Question: Prove that: $\tan \left(\frac{\pi}{4}+x\right)+\tan \left(\frac{\pi}{4}-x\right)=2 \sec 2 x$ Solution: LHS $=\tan \left(\frac{\pi}{4}+x\right)+\tan \left(\frac{\pi}{4}-x\right)$ $=\frac{\tan \frac{\pi}{4}+\tan x}{1-\tan \frac{\pi}{4} \tan x}+\frac{\tan \frac{\pi}{4}-\tan x}{1+\tan \frac{\pi}{4} \tan x} \quad\left[\because \tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B}\right.$ and $\left.\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A \tan B}\right]$ $=\frac{1+\tan x}{1-\tan x}+\frac{1-\tan ...
Read More →Prove that:
Question: Prove that: $\cos ^{6} A-\sin ^{6} A=\cos 2 A\left(1-\frac{1}{4} \sin ^{2} 2 A\right)$ Solution: LHS $=\cos ^{6} A-\sin ^{6} A$ $=\left(\cos ^{2} A\right)^{3}-\left(\sin ^{2} A\right)^{3}$ $=\left(\cos ^{2} A-\sin ^{2} A\right)\left(\cos ^{4} A+\sin ^{2} A \cdot \cos ^{2} A+\sin ^{4} A\right)$ $=\cos 2 A\left(\cos ^{4} A+2 \sin ^{2} A \cos ^{2} A+\sin ^{4} A-\sin ^{2} A \cos ^{2} A\right)$ $=\cos 2 A\left\{\left(\sin ^{2} A+\cos ^{2} A\right)^{2}-\frac{1}{4} \times 4 \sin ^{2} A \cos ^...
Read More →Find the mean of first ten even natural numbers.
Question: Find the mean of first ten even natural numbers. Solution: The first five even natural numbers are 2 , 4 , 6 , 8 , 10 , 12 , 14 , 16 , 18 , 20 $\therefore$ Mean $=\frac{\text { Sum of numbers }}{\text { Total numbers }}$ $=\frac{2+4+6+8+10+12+14+16+18+20}{10}=11$ Mean = 11...
Read More →Find the direction cosines of the sides of the triangle whose vertices are (3, 5, − 4), (− 1, 1, 2) and (− 5, − 5, − 2)
Question: Find the direction cosines of the sides of the triangle whose vertices are (3, 5, 4), ( 1, 1, 2) and ( 5, 5, 2) Solution: The vertices of ΔABC are A (3, 5, 4), B (1, 1, 2), and C (5, 5, 2). The direction ratios of side AB are (1 3), (1 5), and (2 (4)) i.e., 4, 4, and 6. Then, $\sqrt{(-4)^{2}+(-4)^{2}+(6)^{2}}=\sqrt{16+16+36}$ $=\sqrt{68}$ $=2 \sqrt{17}$ Therefore, the direction cosines of AB are $\frac{-4}{\sqrt{(-4)^{2}+(-4)^{2}+(6)^{2}}}, \frac{-4}{\sqrt{(-4)^{2}+(-4)^{2}+(6)^{2}}}, ...
Read More →Show that:
Question: Show that: $2\left(\sin ^{6} x+\cos ^{6} x\right)-3\left(\sin ^{4} x+\cos ^{4} x\right)+1=0$ Solution: LHS $=2\left(\sin ^{6} x+\cos ^{6} x\right)-3\left(\sin ^{4} x+\cos ^{4} x\right)+1$' $=2\left\{\left(\sin ^{2} x\right)^{3}+\left(\cos ^{2}\right)^{3}\right\}-3\left(\sin ^{4} x+\cos ^{4} x\right)+1$ $=2\left[\left(\sin ^{2} x+\cos ^{2} x\right)\left(\sin ^{4} x-\sin ^{2} x \cos ^{2} x+\cos ^{4} x\right)\right]-3\left(\sin ^{4} x+\cos ^{4} x\right)+1$ $=2\left(\sin ^{4} x+\cos ^{4} x...
Read More →Find the mean of all factors of 10.
Question: Find the mean of all factors of 10. Solution: All factors of 6 are 1, 2, 5, 10. $\therefore$ Mean $=\frac{\text { Sum of factors }}{\text { Total factors }}$ $=\frac{1+2+5+10}{4}=4.5$ Mean = 4.5...
Read More →Prove that:
Question: Show that: $3(\sin x-\cos x)^{4}+6(\sin x+\cos )^{2}+4\left(\sin ^{6} x+\cos ^{6} x\right)=13$ Solution: $\mathrm{LHS}=3(\sin x-\cos x)^{4}+6(\sin x+\cos x)^{2}+4\left(\sin ^{6} x+\cos ^{6} x\right)$ $=3\left\{(\sin x-\cos x)^{2}\right\}^{2}+6\left(\sin ^{2} x+\cos ^{2} x+2 \sin x \cos x\right)+4\left\{\left(\sin ^{2} x\right)^{3}+\left(\cos ^{2} x\right)^{3}\right\}$ $=3\left(\sin ^{2} x+\cos ^{2} x-2 \sin x \cos x\right)^{2}+6(1+\sin 2 x)+4\left(\sin ^{2} x+\cos ^{2} x\right)\left(\s...
Read More →Find the mean of first five natural numbers.
Question: Find the mean of first five natural numbers. Solution: The first five odd numbers are 1, 2, 3, 4, 5. $\therefore$ Mean $=\frac{\text { Sum of numbers }}{\text { Total numbers }}$ $=\frac{1+2+3+4+5}{5}$ = 15/5= 3 Mean = 3...
Read More →If sin θ=35, evaluate cos θ−1tan θ2 cot θ.
Question: If $\sin \theta=\frac{3}{5}$, evaluate $\frac{\cos \theta-\frac{1}{\tan \theta}}{2 \cot \theta}$. Solution: Given: $\sin \theta=\frac{3}{5}$.....(1) To find the value of $\frac{\cos \theta-\frac{1}{\tan \theta}}{2 \cot \theta}$ Now, we know the following trigonometric identity $\cos ^{2} \theta+\sin ^{2} \theta=1$ Therefore, by substituting the value of $\sin \theta$ from equation (1), We get, $\cos ^{2} \theta+\left(\frac{3}{5}\right)^{2}=1$ Therefore, $\cos ^{2} \theta=1-\left(\frac{...
Read More →