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- Question 2, Exercise 2.3 @math-11-kpk:sol:unit02
- & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}\right. \right] \\ \underset{\sim}{R} & \left[\begin{matrix} 1 & 4 & 14 \\ 0 & 5 & 6 \\ -1 & 2 & 3 \end{matri... & 0 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{matrix} \right. \right]\text{ by }R_1+3R_3\text{ and }R_2+2R_3 \\ \underset{\sim}{R}&\left[\begin{matrix} 1 & 4 & 14 \
- Question 8 & 9, Review Exercise 10 @math-11-kpk:sol:unit10
- he identity $\sin \left( \dfrac{\pi }{4}-\theta \right)\sin \left( \dfrac{\pi }{4}+\theta \right)=\dfrac{1}{2}\cos 2\theta $. ====Solution==== We know that $2... in \alpha \sin \beta =\cos \left( \alpha -\beta \right)-\cos \left( \alpha +\beta \right)$ \begin{align}L.H.S.&=\sin \left( \dfrac{\pi }{4}-\theta \right)\s
- Question 5, Exercise 10.1 @math-11-kpk:sol:unit10
- uadrant, then find: $\sin \left( \alpha +\beta \right)$. ====Solution==== Given: $\tan\alpha =\dfrac{3... lign}{{\sec}^{2}}\alpha &=1+{{\tan}^{2}}\alpha\\ \Rightarrow \quad \sec \alpha &=\pm \sqrt{1+{{\tan}^{2}}... gn}\sec \alpha &=-\sqrt{1+{{\tan}^{2}}\alpha }\\ \Rightarrow \quad \sec \alpha &=-\sqrt{1+{{\left(\frac{3}{4} \right)}^{2}}}=-\sqrt{1+\frac{9}{16}}=-\sqrt{\frac{25}{1
- Question 3, Exercise 2.1 @math-11-kpk:sol:unit02
- ix}x\\y\\z\end{bmatrix}$. Verify that $\left( AB \right)C=A\left( BC \right)$. ====Solution==== Given: $A=\begin{bmatrix}x & y & z\end{bmatrix}$, $B=\begin{bmat... \left[ x( ax+hy+gz )+y( hx+by+fz )+z( gx+fy+cz ) \right] \\ &=\left[ a{{x}^{2}}+hxy+gxz+hxy+b{{y}^{2}}+fyz+gxz+fyz+c{{z}^{2}} \right] \\ &=\left[ a{{x}^{2}}+2hxy+2gxz+b{{y}^{2}}+2fyz
- Question 2, Exercise 10.1 @math-11-kpk:sol:unit10
- eta -\cos \alpha \sin.\end{align} \begin{align} \Rightarrow \quad \sin \left( \frac{\pi }{3}-\frac{\pi }{4} \right) & =\sin \frac{\pi }{3}\cos \frac{\pi }{4}-\cos \... in \frac{\pi }{4} \\ &=\left( \frac{\sqrt{3}}{2} \right)\left( \frac{1}{\sqrt{2}} \right)-\left( \frac{1}{2} \right)\left( \frac{1}{\sqrt{2}} \right) \\ & =\fra
- Question 12 & 13, Exercise 3.3 @math-11-kpk:sol:unit03
- n 12===== Prove that the angle in a semicircle is right angle. ====Solution==== We are considering a tria... side a semicircle as shown. We have to show $\overrightarrow{B A} \cdot \overrightarrow{A C}=0$. We see in figure that: $|\vec{a}|=\vec{b}|=| \vec{c} \mid=$ ra... From $\triangle A B O$, we have \begin{align}\overrightarrow{O B}+\overrightarrow{A B}&=\overrightarrow{O
- Question 5 Exercise 3.4 @math-11-kpk:sol:unit03
- {align}\text{Area of triangle}&=\dfrac{1}{2}|\overrightarrow{P Q} \times \overrightarrow{P R}| \\ \text { Since } \overrightarrow{P Q}&=(3,2)-(-2 ,-3) \\ \Rightarrow \overrightarrow{P Q}&=(5,5) \\ \overrightarrow{P R}&=(-1,-8)-(-
- Question 5, Exercise 10.3 @math-11-kpk:sol:unit10
- os \alpha \cos \beta =\cos \left( \alpha +\beta \right)+\cos \left( \alpha -\beta \right)$\\ \begin{align}L.H.S.&=\cos {{20}^{\circ }}\cos {{40}^{\circ }}\cos... {\circ }}\cos {{40}^{\circ }}\left( \dfrac{1}{2} \right)\cos {{80}^{\circ }}\\ &=\dfrac{1}{2}\cos {{80}^{... eft( 2\,\cos {{80}^{\circ }}\cos {{40}^{\circ }} \right)\cos {{20}^{\circ }}\\ &=\dfrac{1}{4}\left( \cos
- Question 5, Exercise 10.3 @math-11-kpk:sol:unit10
- os \alpha \cos \beta =\cos \left( \alpha +\beta \right)+\cos \left( \alpha -\beta \right)$\\ \begin{align}L.H.S.&=\cos {{20}^{\circ }}\cos {{40}^{\circ }}\cos... {\circ }}\cos {{40}^{\circ }}\left( \dfrac{1}{2} \right)\cos {{80}^{\circ }}\\ &=\dfrac{1}{2}\cos {{80}^{... eft( 2\,\cos {{80}^{\circ }}\cos {{40}^{\circ }} \right)\cos {{20}^{\circ }}\\ &=\dfrac{1}{4}\left( \cos
- Question 1, Exercise 2.1 @math-11-kpk:sol:unit02
- \left[ \begin{matrix} 1 & 2 & 4 \\ \end{matrix} \right] \left[ \begin{matrix} 1 & 0 & 2 \\ 2 & 0 & 1 \\ 0 & 1 & 2 \\ \end{matrix} \right] \left[ \begin{matrix} 2 \\ 4 \\ 6 \\ \end{matrix} \right]$$ ====Solution==== \begin{align}&\left[ \begin{matrix} 1 & 2 & 4 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 0 & 2 \\ 2 & 0 &
- Question 10 Exercise 7.2 @math-11-kpk:sol:unit07
- ^n=\left(\begin{array}{l} n \\ \vdots \end{array}\right)+\left(\begin{array}{l} m \\ 1 \end{array}\right) x+\left(\begin{array}{l} n \\ 2 \end{array}\right) x^2-\ldots+i_n^*\right) x^n \cdot $$ Putting $x=1$ in the above equation, we have $(1 \div 1)^n=\left(
- Question 9 & 10, Exercise 3.2 @math-11-kpk:sol:unit03
- eshawar, Pakistan. =====Question 9===== If $\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k}, $$\overrightarrow{b}=4\hat{i}-2\hat{j}+3\hat{k}$ and $\overrightarrow{c}=\hat{i}-2\hat{j}+\hat{k}$, find a vector of ... f $6$ unit which is parallel to the vector $2\overrightarrow{a}-\overrightarrow{b}+3\overrightarrow{c}.$
- Question 3 & 4, Exercise 1.2 @math-11-kpk:sol:unit01
- \begin{align}{{z}_{1}}\left( {{z}_{2}}+{{z}_{3}} \right)&={{z}_{1}}{{z}_{2}}+{{z}_{1}}{{z}_{3}}\\ {{z}_{... }&=\sqrt{2}-\sqrt{3}i+2+3i\\ &=\left( \sqrt{2}+2 \right)+\left( 3-\sqrt{3} \right)i\\ L.H.S.&={{z}_{1}}\left( {{z}_{2}}+{{z}_{3}} \right)\\ &=\left( \sqrt{3}+\sqrt{2}i \right)\left( \lef
- Question 3, Exercise 10.1 @math-11-kpk:sol:unit10
- e each of the following exactly $\cos \left( u+v \right)$ ====Solution==== Given $\sin u=\dfrac{3}{5},$ ... \cos $ is $+ve$ in first quadrant . \begin{align}\Rightarrow \,\,\,\,\,\cos u&=\sqrt{1-{{\sin }^{2}}u}\\ &=\sqrt{1-{{\left( \frac{3}{5} \right)}^{2}}}\\ &=\sqrt{1-\dfrac{9}{25}}\,\,=\sqrt{\dfrac{16}{25}}\\ \Rightarrow \,\,\,\,\,\cos u&=\dfrac{4}{5}\end{align} Al
- Question 11, Exercise 3.3 @math-11-kpk:sol:unit03
- \hat{k}$ and $2 \hat{i}+\hat{j}-4 \hat{k}$ form a right angle triangle. ====Solution==== Let $\vec{a}=3 \... gin{align}|\vec{a}|&=\sqrt{(3)^2+(-2)^2+(1)^2}\\ \Rightarrow \quad |\vec{a}|&=\sqrt{14},\\ |\vec{b}|&=\sq... c}$ represent the sides of triangle and they form right angle triangle. Also if we see \begin{align}\vec{... }+\hat{k}) \cdot(2 \hat{i}+\hat{j}-4 \hat{k}) \\ \Rightarrow \vec{a} \cdot \vec{c}&=6 - 2-4=0 .\end{align