Find the angle between two vectors
Question: Find the angle between two vectors $\vec{a}$ and $\vec{b}$ with magnitudes $\sqrt{3}$ and 2 , respectively having $\vec{a} \cdot \vec{b}=\sqrt{6}$. Solution: It is given that, $|\vec{a}|=\sqrt{3},|\vec{b}|=2$ and, $\vec{a} \cdot \vec{b}=\sqrt{6}$ Now, we know that $\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta$. $\therefore \sqrt{6}=\sqrt{3} \times 2 \times \cos \theta$ $\Rightarrow \cos \theta=\frac{\sqrt{6}}{\sqrt{3} \times 2}$ $\Rightarrow \cos \theta=\frac{1}{\sqrt{2}}$ $\Ri...
Read More →If cos (A + B) sin (C − D) = cos (A − B) sin (C + D),
Question: If cos (A+B) sin (CD) = cos (AB) sin (C+D), prove that tanAtanBtanC+ tanD= 0. Solution: cos (A+B) sin (CD) = cos (AB) sin (C+D) $\Rightarrow[\cos A \cos B-\sin A \sin B][\sin C \cos D-\cos C \sin D]=[\cos A \cos B+\sin A \sin B][\sin C \cos D+\cos C \sin D]$ Dividing both sides by $\cos A \cos B \cos C \cos D$, $\frac{[\cos A \cos B-\sin A \sin B][\sin C \cos D-\cos C \sin D]}{\cos A \cos B \cos C \cos D}=\frac{[\cos A \cos B+\sin A \sin B][\sin C \cos D+\cos C \sin D]}{\cos A \cos B \...
Read More →If cos (A + B) sin (C − D) = cos (A − B) sin (C + D),
Question: If cos (A+B) sin (CD) = cos (AB) sin (C+D), prove that tanAtanBtanC+ tanD= 0. Solution: cos (A+B) sin (CD) = cos (AB) sin (C+D) $\Rightarrow[\cos A \cos B-\sin A \sin B][\sin C \cos D-\cos C \sin D]=[\cos A \cos B+\sin A \sin B][\sin C \cos D+\cos C \sin D]$ Dividing both sides by $\cos A \cos B \cos C \cos D$, $\frac{[\cos A \cos B-\sin A \sin B][\sin C \cos D-\cos C \sin D]}{\cos A \cos B \cos C \cos D}=\frac{[\cos A \cos B+\sin A \sin B][\sin C \cos D+\cos C \sin D]}{\cos A \cos B \...
Read More →If are two collinear vectors, then which of the following are incorrect:
Question: If $\vec{a}$ and $\vec{b}$ are two collinear vectors, then which of the following are incorrect: A. $\vec{b}=\lambda \vec{a}$, for some scalar $\lambda$ B. $\vec{a}=\pm \vec{b}$ C. the respective components of $\vec{a}$ and $\vec{b}$ are proportional D. both the vectors $\vec{a}$ and $\vec{b}$ have same direction, but different magnitudes Solution: If $\vec{a}$ and $\vec{b}$ are two collinear vectors, then they are parallel. Therefore, we have: $\vec{b}=\lambda \vec{a}$ (For some scala...
Read More →If y sin ϕ = x sin (2θ + ϕ),
Question: Ifysin ϕ =xsin (2 + ϕ), prove that (x+y) cot ( + ϕ) = (yx) cot . Solution: Given: ysin ϕ =xsin (2 + ϕ) $\Rightarrow \frac{y}{x}=\frac{\sin (2 \theta+\phi)}{\sin \phi}$ Applying componendo and dividendo : $\Rightarrow \frac{y-x}{y+x}=\frac{\sin (2 \theta+\phi)-\sin \phi}{\sin (2 \theta+\phi)+\sin \phi}$ $\Rightarrow \frac{y-x}{y+x} \Rightarrow \frac{2 \sin \left(\frac{2 \theta+\phi-\phi}{2}\right) \cos \left(\frac{2 \theta+\phi+\phi}{2}\right)}{2 \sin \left(\frac{2 \theta+\phi+\phi}{2}\...
Read More →If y sin ϕ = x sin (2θ + ϕ),
Question: Ifysin ϕ =xsin (2 + ϕ), prove that (x+y) cot ( + ϕ) = (yx) cot . Solution: Given: ysin ϕ =xsin (2 + ϕ) $\Rightarrow \frac{y}{x}=\frac{\sin (2 \theta+\phi)}{\sin \phi}$ Applying componendo and dividendo : $\Rightarrow \frac{y-x}{y+x}=\frac{\sin (2 \theta+\phi)-\sin \phi}{\sin (2 \theta+\phi)+\sin \phi}$ $\Rightarrow \frac{y-x}{y+x} \Rightarrow \frac{2 \sin \left(\frac{2 \theta+\phi-\phi}{2}\right) \cos \left(\frac{2 \theta+\phi+\phi}{2}\right)}{2 \sin \left(\frac{2 \theta+\phi+\phi}{2}\...
Read More →In a ∆ABC, right angled at B, AB = 24 cm, BC = 7 cm. Determine
Question: In a ∆ABC, right angled at B, AB = 24 cm, BC = 7 cm. Determine (i) sin A, cos A(ii) sin C, cos C Solution: (i) The given triangle is below:- $B C=7 \mathrm{~cm}$ $\angle A B C=90^{\circ}$ To Find: $\sin A, \cos A$ In this problem, Hypotenuse side is unknown Hence we first find Hypotenuse side by Pythagoras theorem By Pythagoras theorem, We get, $A C^{2}=A B^{2}+B C^{2}$ $A C^{2}=24^{2}+7^{2}$ $A C^{2}=576+49$ $A C^{2}=625$ $A C=\sqrt{625}$ $A C=25$ Hypotenuse $=25$ By definition, $\sin...
Read More →If cos (α + β) sin (γ + δ) = cos (α − β) sin (γ − δ),
Question: If cos ( + ) sin ( + ) = cos ( ) sin ( ), prove that cot cot cot = cot Solution: $\cos (\alpha+\beta) \sin (\gamma+\delta)=\cos (\alpha-\beta) \sin (\gamma-\delta)$ $\Rightarrow[\cos \alpha \cos \beta-\sin \alpha \sin \beta][\sin \gamma \cos \delta+\cos \gamma \sin \delta]=$ $[\cos \alpha \cos \beta+\sin \alpha \sin \beta][\sin \gamma \cos \delta-\cos \gamma \sin \delta]$ Dividing both sides by $\sin \alpha \sin \beta \sin \gamma \sin \delta:$ $\frac{[\cos \alpha \cos \beta-\sin \alpha...
Read More →If
Question: If $\frac{\cos (A-B)}{\cos (A+B)}+\frac{\cos (C+D)}{\cos (C-D)}=0$, Prove that $\tan A \tan B \tan C \tan D=-1$ Solution: We have, $\frac{\cos (A-B)}{\cos (A+B)}+\frac{\cos (C+D)}{\cos (C-D)}=0$ $\Rightarrow \frac{\cos (A-B) \cos (C-D)+\cos (C+D) \cos (A+B)}{\cos (A+B) \cos (C-D)}=0$ $\Rightarrow \cos (A-B) \cos (C-D)+\cos (C+D) \cos (A+B)=0$ $\Rightarrow \cos (A-B) \cos (C-D)=-\cos (C+D) \cos (A+B)$ $\Rightarrow[\cos A \cos B+\sin A \sin B][\cos C \cos D+\sin C \sin D]=$$-[\cos C \cos...
Read More →In triangle ABC which of the following is not true:
Question: In triangle ABC which of the following isnottrue: A. $\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}+\overrightarrow{\mathrm{CA}}=\overrightarrow{0}$ B. $\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}-\overrightarrow{\mathrm{AC}}=\overrightarrow{0}$ C. $\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}-\overrightarrow{\mathrm{CA}}=\overrightarrow{0}$ D. $\overrightarrow{\mathrm{AB}}-\overrightarrow{\mathrm{CB}}+\overrightarrow{\mathrm{CA}}=\overrightarrow{...
Read More →Prove that:
Question: Prove that: (i) $\frac{\cos (A+B+C)+\cos (-A+B+C)+\cos (A-B+C)+\cos (A+B-C)}{\sin (A+B+C)+\sin (-A+B+C)+\sin (A-B+C)-\sin (A+B-C)}=\cot C$ (ii) $\sin (B-C) \cos (A-D)+\sin (C-A) \cos (B-D)+\sin (A-B) \cos (C-D)=0$ Solution: (i) Goncidor LHS. $\frac{\cos (A+B+C)+\cos (-A+B+C)+\cos (A-B+C)+\cos (A+B-C)}{\sin (A+B+C)+\sin (-A+B+C)+\sin (A-B+C)-\sin (A+B-C)}$ $=\frac{2 \cos \left(\frac{\mathrm{A}+\mathrm{B}+\mathrm{C}-\mathrm{A}+\mathrm{B}+\mathrm{C}}{2}\right) \cos \left(\frac{\mathrm{A}+...
Read More →The diameter of the moon is approximately one-fourth of the diameter of the earth.
Question: The diameter of the moon is approximately one-fourth of the diameter of the earth. What is the earth the volume of the moon? Solution: Diameter of moon = 1/4th diameter of earth Let the diameter of earth be d, so radius = d/2 Then diameter of moon = d/4 Radius $=\frac{\frac{\mathrm{d}}{2}}{4}=\frac{\mathrm{d}}{8}$ Volume of moon $=\frac{4}{3} \pi r^{3}=\frac{4}{3} \pi\left(\frac{d}{8}\right)^{3}=\frac{4}{3} \times \frac{1}{512} \pi d^{3}$ Volume of earth $=\frac{4}{3} \pi r^{3}=\frac{4...
Read More →A capsule of medicine is in the shape of a sphere of diameter 3.5 mm.
Question: A capsule of medicine is in the shape of a sphere of diameter $3.5 \mathrm{~mm}$. How much medicine $\left(\mathrm{mm}^{3}\right)$ is needed to fill this capsule? Solution: Given that Diameter of capsule = 3.5 mm Radius = 3.5/2 = 1.75 mm Volume of spherical sphere $=4 / 3 \pi r^{3}$ $=4 / 3 \times 22 / 7 \times(1.75)^{3}$ $=22.458 \mathrm{~mm}^{3}$ Therefore $22.46 \mathrm{~mm}^{3}$ of medicine is required...
Read More →Show that the points A, B and C with position vectors,
Question: Show that the points $A, B$ and $C$ with position vectors, $\vec{a}=3 \hat{i}-4 \hat{j}-4 \hat{k}, \vec{b}=2 \hat{i}-\hat{j}+\hat{k}$ and $\vec{c}=\hat{i}-3 \hat{j}-5 \hat{k}$, respectively form the vertices of a right angled triangle. Solution: Position vectors of points A, B, and C are respectively given as: $\vec{a}=3 \hat{i}-4 \hat{j}-4 \hat{k}, \vec{b}=2 \hat{i}-\hat{j}+\hat{k}$ and $\vec{c}=\hat{i}-3 \hat{j}-5 \hat{k}$ $\vec{a}=3 \hat{i}-4 \hat{j}-4 \hat{k}, \vec{b}=2 \hat{i}-\ha...
Read More →A hemispherical tank is made up of an iron sheet 1 cm thick.
Question: A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1 m, then find the volume of the iron used to make the tank. Solution: Given that, Inner radius of the hemispherical tank = 1 m = r1 Thickness of the hemispherical tank = 1 cm = 0.01 m Outer radius of hemispherical tank = (1 + 0.01) = 1.01 m = r2 Volume of iron used to make the tank $=\frac{2}{3} \pi\left(r_{2}^{3}-r_{1}^{3}\right)$ $=\frac{2}{3} \times \frac{22}{7}\left[(1.01)^{3}-1^{3}\right]$ $=\frac...
Read More →If sin 2A=λ sin 2B,
Question: If $\sin 2 A=\lambda \sin 2 B$, prove that $\frac{\tan (A+B)}{\tan (A-B)}=\frac{\lambda+1}{\lambda-1}$. Solution: Given: sin 2A= sin 2B $\Rightarrow \frac{\sin 2 A}{\sin 2 B}=\lambda$ $\Rightarrow \frac{\sin 2 A+\sin 2 B}{\sin 2 A-\sin 2 B}=\frac{\lambda+1}{\lambda-1}$ $\Rightarrow \frac{2 \sin \left(\frac{2 A+2 B}{2}\right) \cos \left(\frac{2 A-2 B}{2}\right)}{2 \sin \left(\frac{2 A-2 B}{2}\right) \cos \left(\frac{2 A+2 B}{2}\right)}=\frac{\lambda+1}{\lambda-1}$ $\left[\because \sin A...
Read More →A cube of side 4 cm contains a sphere touching its side.
Question: A cube of side 4 cm contains a sphere touching its side. Find the volume of the gap in between. Solution: It is given that Cube side = 4cm Volume of cube $=(4 \mathrm{~cm})^{3}=64 \mathrm{~cm}^{3}$ Diameter of the sphere = Length of the side of the cube = 4cm Therefore radius of the sphere = 2cm Volume of the sphere $=4 / 3 \pi r^{3}=4 / 3 \times 22 / 7 \times(2)^{3}=33.52 \mathrm{~cm}^{3}$ Volume of gap = Volume of cube - Volume of sphere $=64 \mathrm{~cm}^{3}-33.52 \mathrm{~cm}^{3}=3...
Read More →Find the position vector of the mid point of the vector joining the points P (2, 3, 4) and Q (4, 1, – 2).
Question: Find the position vector of the mid point of the vector joining the points P (2, 3, 4) and Q (4, 1, 2). Solution: The position vector of mid-point R of the vector joining points P (2, 3, 4) and Q (4, 1, 2) is given by, $\overrightarrow{\mathrm{OR}}=\frac{(2 \hat{i}+3 \hat{j}+4 \hat{k})+(4 \hat{i}+\hat{j}-2 \hat{k})}{2}=\frac{(2+4) \hat{i}+(3+1) \hat{j}+(4-2) \hat{k}}{2}$ $=\frac{6 \hat{i}+4 \hat{j}+2 \hat{k}}{2}=3 \hat{i}+2 \hat{j}+\hat{k}$...
Read More →Find the position vector of a point
Question: Find the position vector of a point $\mathrm{R}$ which divides the line joining two points $\mathrm{P}$ and $\mathrm{Q}$ whose position vectors are $\hat{i}+2 \hat{j}-\hat{k}$ and $-\hat{i}+\hat{j}+\hat{k}$ respectively, in the ration $2: 1$ (i) internally (ii) externally Solution: The position vector of point R dividing the line segment joining two points P and Q in the ratiom:nis given by: i. Internally: $\frac{m \vec{b}+n \vec{a}}{m+n}$ ii. Externally: $\frac{m \vec{b}-n \vec{a}}{m-...
Read More →A sphere, a cylinder, and a cone have the same diameter.
Question: A sphere, a cylinder, and a cone have the same diameter. The height of the cylinder and also the cone are equal to the diameter of the sphere. Find the ratio of their volumes. Solution: Let r be the common radius Height of the cone = height of the cylinder = 2r Let $y_{1}=$ Volume of sphere $=4 / 3 \pi r^{3}$ $\mathrm{v}_{1}=$ Volume of cylinder $=\pi r^{2} \mathrm{~h}=\pi r^{2} \times 2 r$ $v_{1}=$ Volume of cone $=1 / 3 \pi r^{2} h=1 / 3 \pi r^{3}$ Now $v_{1}: v_{2}: v_{3}=4 / 3 \pi ...
Read More →If cosec A + sec A = cosec B + sec B,
Question: If $\operatorname{cosec} A+\sec A=\operatorname{cosec} B+\sec B$, prove that $\tan A \tan B=\cot \frac{A+B}{2}$. Solution: Given: $\frac{1}{\sin A}+\frac{1}{\cos A}=\frac{1}{\sin B}+\frac{1}{\cos B}$ $\Rightarrow \frac{1}{\sin A}-\frac{1}{\sin B}=\frac{1}{\cos B}-\frac{1}{\cos A}$ $\Rightarrow \frac{\sin B-\sin A}{\sin A \sin B}=\frac{\cos A-\cos B}{\cos A \cos B}$ $\Rightarrow \frac{\sin B-\sin A}{\cos A-\cos B}=\frac{\sin A \sin B}{\cos A \cos B}$ $\Rightarrow \frac{2 \sin \left(\fra...
Read More →In each of the following, one of the six trigonometric ratios is given.
Question: In each of the following, one of the six trigonometric ratios is given. Find the values of the other trigonometric ratios. (i) $\sin A=\frac{2}{3}$ (ii) $\cos A=\frac{4}{5}$ (iii) $\tan \theta=11$ (iv) $\sin \theta=\frac{11}{15}$ (v) $\tan \alpha=\frac{5}{12}$ (vi) $\sin \theta=\frac{\sqrt{3}}{2}$ (vii) $\cos \theta=\frac{7}{25}$ (viii) $\tan \theta=\frac{8}{15}$ (ix) $\cot \theta=\frac{12}{5}$ (x) $\sec \theta=\frac{13}{5}$ (xi) $\operatorname{cosec} \theta=\sqrt{10}$ (xii) $\cos \the...
Read More →The largest sphere is carved out of a cube of side 10.5 cm.
Question: The largest sphere is carved out of a cube of side 10.5 cm. Find the volume of the sphere. Solution: Side of cube = 10.5 cm Volume of sphere = v Diameter of the largest sphere = 10.5 cm 2r = 10.5 Volume of sphere $=4 / 3 \pi r^{3}=4 / 3 \times 22 / 7 \times 5.25 \times 5.25 \times 5.25$ $\mathrm{v}=\frac{11 \times 441}{8} \mathrm{~cm}^{3}$ $v=606.375 \mathrm{~cm}^{3}$ r = 5.25 cm...
Read More →Show that the vector
Question: Show that the vector $\hat{i}+\hat{j}+\hat{k}$ is equally inclined to the axes $O X, O Y$, and $O Z$. Solution: Let $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ Then, $|\vec{a}|=\sqrt{1^{2}+1^{2}+1^{2}}=\sqrt{3}$ Therefore, the direction cosines of $\vec{a}$ are $\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)$ Now, let $\alpha, \beta$, and $\gamma$ be the angles formed by $\vec{a}$ with the positive directions of $x, y$, and $z$ axes. Then, we have $\cos \alpha=\frac{1}{\...
Read More →If
Question: If $\cos A+\cos B=\frac{1}{2}$ and $\sin A+\sin B=\frac{1}{4}$, prove that $\tan \left(\frac{A+B}{2}\right)=\frac{1}{2}$. Solution: Given: $\sin A+\sin B=\frac{1}{4}$ ....(i) $\cos A+\cos B=\frac{1}{2}$ ....(ii) Dividing (i) by (ii): $\Rightarrow \frac{\sin A+\sin B}{\cos A+\cos B}=\frac{\frac{1}{4}}{\frac{1}{2}}$ $\Rightarrow \frac{2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)}{2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)}=\frac{1}{2}$ $\left[\...
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