第12类RD Sharma解决方案–第23章，矢量代数–练习23.9

问题1：矢量的方向角可以为45°，60°和120°吗?

解决方案：

We know that if l, m and n are the direction cosines and $\alpha$, $\beta$ and $\gamma$ are the direction angles then,

=> $l = \cos \alpha = \cos 45 \degree = \dfrac{1}{\sqrt{2}}$

=> $m = \cos \beta = \cos 60 \degree = \dfrac{1}{2}$

=> $n = \cos \gamma = \cos 120 \degree = \dfrac{1}{2}$

Also,

=> l² + m² + n² = 1

=> $(\dfrac{1}{\sqrt{2}})^2 + (\dfrac{1}{2})^2+ (\dfrac{1}{2})^2 = 1$

=> $\dfrac{1}{2}+\dfrac{1}{4} +\dfrac{1}{4} = 1$

=> As LHS = RHS, the vector can have these direction angles.

问题2：证明1,1和1不能为直线的方向余弦。

解决方案：

Given that, l=1, m=1 and n=1.

We know that,

=> l² + m² + n² = 1

=> 1² + 1² + 1² = 1

=> 3 ≠ 1

Thus, 1, 1 and 1 can never be the direction cosines of a straight line.

=> Hence proved.

问题3：向量成一个角度$\dfrac{\pi}{4}$分别具有x轴和y轴。找到它与z轴所成的角度。

解决方案：

We know that if l, m and n are the direction cosines and $\alpha$, $\beta$ and $\gamma$ are the direction angles then,

=> $l = \cos \alpha = \cos 45 \degree = \dfrac{1}{\sqrt{2}}$

=> $m = \cos \beta = \cos 45 \degree = \dfrac{1}{\sqrt{2}}$

Let $\gamma$ be the angle we have to calculate.

We know that,

=> l² + m² + n² = 1

=> $n^2 = 1- [(\dfrac{1}{\sqrt{2}})^2+(\dfrac{1}{\sqrt{2}})^2]$

=> n² = 1 – 1

=> n² = 0

=> $\cos^2 \gamma = 0$

=> $\cos \gamma = 0$

=> $\gamma = \cos^{-1}0$

=> $\gamma = \dfrac{\pi}{2}$

问题4：向量$\vec{r}$相对于x轴，y轴和z轴以相等的锐角倾斜。如果$|\vec{r}|$ = 6个单位，找到$\vec{r}$ 。

解决方案：

Given that $\alpha=\beta=\gamma$

=> $\cos \alpha = \cos \beta = \cos \gamma$

=> l = m = n = p (say)

We know that,

=> l² + m² + n² = 1

=> p² + p² + p² = 1

=> 3p² = 1

=> $p = \pm \dfrac{1}{\sqrt{3}}$

The vector $\vec{r}$ can be described as,

=> $\vec{r} = |\vec{r}|(l\hat{i}+m\hat{j}+n\hat{k})$

=> $\vec{r} = 6(\pm \dfrac{1}{\sqrt{3}} \hat{i}+ \pm \dfrac{1}{\sqrt{3}}\hat{j} +\pm \dfrac{1}{\sqrt{3}}\hat{k})$

=>$\vec{r} = \pm2\sqrt{3}(\hat{i}+\hat{j}+\hat{k})$

问题5：向量$\vec{r}$相对于x轴倾斜45°，而y轴倾斜60°。如果$|\vec{r}|=8$单位，找到$\vec{r}$ 。

解决方案：

Given that $\alpha = 45 \degree$ and $\beta = 60 \degree$

We know that,

=> l² + m² + n² = 1

=> $\cos ^2 \alpha + \cos ^2 \beta + \cos^2 \gamma = 1$

=> $(\dfrac{1}{\sqrt{2}})^2+ (\dfrac{1}{2})^2 + \cos^2 \gamma = 1$

=> $\cos^2 \gamma = 1- [\dfrac{1}{2}+\dfrac{1}{4}]$

=> $\cos ^2 \gamma = 1- \dfrac{3}{4}$

=> $\cos ^2 \gamma = \dfrac{1}{4}$

=> $\cos \gamma = \pm \dfrac{1}{2}$

The vector $\vec{r}$ can be described as,

=> $\vec{r} = |\vec{r}|(l\hat{i}+m\hat{j}+n\hat{k})$

=> $\vec{r} = 8(\dfrac{1}{\sqrt{2}} \hat{i}+ \dfrac{1}{2}\hat{j} +\pm \dfrac{1}{2}\hat{k})$

=> $\vec{r} = 4(\sqrt{2}\hat{i}+\hat{j}+\pm \hat{k})$

问题6：找到以下向量的方向余弦：

(一世)： $2\hat{i}+ 2\hat{j}-\hat{k}$

解决方案：

The direction ratios are given as 2, 2 and -1.

Direction cosines are given as,

=> $\cos \alpha, \cos \beta, \cos \gamma = \dfrac{2}{|\vec{r}|}, \dfrac{2}{|\vec{r}|}, \dfrac{-1}{|\vec{r}|}$

=> $\cos \alpha, \cos \beta , \cos \gamma = \dfrac{2}{\sqrt{2^2+2^2+(-1)^2}}, \dfrac{2}{\sqrt{2^2+2^2+(-1)^2}}, \dfrac{-1}{\sqrt{2^2+2^2+(-1)^2}}$

=> $\cos \alpha, \cos \beta , \cos \gamma = \dfrac{2}{3}, \dfrac{2}{3}, \dfrac{-1}{3}$

(ii)： $6\hat{i}-2\hat{j}-3\hat{k}$

解决方案：

The direction ratios are given as 6, -2 and -3.

Direction cosines are given as,

=> $\cos \alpha, \cos \beta , \cos \gamma = \dfrac{6}{|\vec{r}|}, \dfrac{-2}{|\vec{r}|}, \dfrac{-3}{|\vec{r}|}$

=> $\cos \alpha, \cos \beta , \cos \gamma = \dfrac{6}{\sqrt{6^2+(-2)^2+(-3)^2}}, \dfrac{-2}{\sqrt{6^2+(-2)^2+(-3)^2}}, \dfrac{-3}{\sqrt{6^2+(-2)^2+(-3)^2}}$

=> $\cos \alpha, \cos \beta , \cos \gamma = \dfrac{6}{7}, \dfrac{-2}{7}, \dfrac{-3}{7}$

(iii)： $3\hat{i}-4\hat{k}$

解决方案：

The direction ratios are given as 3, 0 and -4.

Direction cosines are given as,

=> $\cos \alpha, \cos \beta , \cos \gamma = \dfrac{3}{|\vec{r}|}, \dfrac{0}{|\vec{r}|}, \dfrac{-4}{|\vec{r}|}$

=> $\cos \alpha, \cos \beta , \cos \gamma = \dfrac{3}{\sqrt{3^2+0^2+(-4)^2}}, \dfrac{0}{\sqrt{3^2+0^2+(-4)^2}}, \dfrac{-4}{\sqrt{3^2+0^2+(-4)^2}}$

=> $\cos \alpha, \cos \beta , \cos \gamma = \dfrac{3}{5}, 0 , \dfrac{-4}{5}$

问题7：找出以下向量相对于每个坐标轴的倾斜角度。

(一世)： $\hat{i}-\hat{j}+\hat{k}$

解决方案：

The given direction ratios are: 1,-1,1.

Thus,

=> $l, m, n = \dfrac{1}{|\vec{r}|}, \dfrac{-1}{|\vec{r}|}, \dfrac{1}{|\vec{r}|}$

=> $l, m, n = \dfrac{1}{\sqrt{1^2+(-1)^2+1^2}}, \dfrac{-1}{\sqrt{1^2+(-1)^2+1^2}}, \dfrac{1}{\sqrt{1^2+(-1)^2+1^2}}$

=> $l, m, n = \dfrac{1}{\sqrt{3}}, \dfrac{-1}{\sqrt{3}} , \dfrac{1}{\sqrt{3}}$

=> $\cos \alpha , \cos \beta , \cos \gamma = \dfrac{1}{\sqrt{3}}, \dfrac{-1}{\sqrt{3}} , \dfrac{1}{\sqrt{3}}$

=> $\alpha , \beta , \gamma = \cos ^{-1} (\dfrac{1}{\sqrt{3}}), \cos ^{-1} (\dfrac{-1}{\sqrt{3}}), \cos ^{-1} (\dfrac{1}{\sqrt{3}})$

(ii)： $\hat{j}-\hat{k}$

解决方案：

The given direction ratios are: 0,1,-1.

Thus,

=> $l, m, n = \dfrac{0}{|\vec{r}|}, \dfrac{1}{|\vec{r}|}, \dfrac{-1}{|\vec{r}|}$

=> $l, m, n = \dfrac{0}{\sqrt{0^2+1^2+(-1)^2}}, \dfrac{1}{\sqrt{0^2+1^2+(-1)^2}}, \dfrac{-1}{\sqrt{0^2+1^2+(-1)^2}}$

=> $l, m, n = 0 , \dfrac{1}{\sqrt{2}} , \dfrac{-1}{\sqrt{2}}$

=> $\cos \alpha , \cos \beta , \cos \gamma = 0, \dfrac{1}{\sqrt{2}} , \dfrac{-1}{\sqrt{2}}$

=> $\alpha , \beta , \gamma = \cos ^{-1}(0), \cos ^{-1} (\dfrac{1}{\sqrt{2}}), \cos ^{-1} (\dfrac{-1}{\sqrt{2}})$

=> $\alpha , \beta , \gamma = \dfrac{\pi}{2}, \dfrac{\pi}{4}, \dfrac{3\pi}{4}$

(iii)： $4\hat{i}+8\hat{j}+\hat{k}$

解决方案：

The given direction ratios are: 4, 8, 1.

Thus,

=> $l, m, n = \dfrac{4}{|\vec{r}|}, \dfrac{8}{|\vec{r}|}, \dfrac{1}{|\vec{r}|}$

=> $l, m, n = \dfrac{0}{\sqrt{4^2+8^2+1^2}}, \dfrac{8}{\sqrt{4^2+8^2+1^2}}, \dfrac{1}{\sqrt{4^2+8^2+1^2}}$

=> $l, m, n = \dfrac{4}{9}, \dfrac{8}{9} , \dfrac{1}{9}$

=> $\cos \alpha , \cos \beta , \cos \gamma = \dfrac{4}{9}, \dfrac{8}{9} , \dfrac{1}{9}$

=> $\alpha , \beta , \gamma = \cos ^{-1}(\dfrac{4}{9}), \cos ^{-1} (\dfrac{8}{9}), \cos ^{-1} (\dfrac{1}{9})$

问题8：证明向量$\hat{i}+\hat{j}+\hat{k}$分别与OX，OY和OZ轴倾斜。

解决方案：

Let $\vec{r}= \hat{i}+\hat{j}+\hat{k}$

Thus, $|\vec{r}| = \sqrt{1^2+1^2+1^2}$

=> $|\vec{r}| = \sqrt{3}$

Thus the direction cosines are: $\dfrac{1}{|\vec{r}|}$, $\dfrac{1}{|\vec{r}|}$ and $\dfrac{1}{|\vec{r}|}$

=> $\cos \alpha, \cos \beta , \cos \gamma = \dfrac{1}{\sqrt{3}}, \dfrac{1}{\sqrt{3}}, \dfrac{1}{\sqrt{3}}$

Thus,

=>$\cos \alpha, \cos \beta , \cos \gamma = \dfrac{1}{\sqrt{3}}$

=> Thus, the vector is equally inclined with the 3 axes.

问题9：证明向量相对于OX，OY和OZ轴等向倾斜的方向余弦为$\dfrac{1}{\sqrt{3}}$ ， $\dfrac{1}{\sqrt{3}}$ ， $\dfrac{1}{\sqrt{3}}$ 。

解决方案：

Let the vector be equally inclined at an angle of $\alpha$.

Then the direction cosines of the vector l, m, n are: $\cos \alpha$, $\cos \beta$ and $\cos \gamma$

We know that,

=> l² + m² + n² = 1

=> $\cos^2 \alpha + \cos^2 \alpha+ \cos^2 \alpha = 1$

=> $3\cos^2 \alpha = 1$

=> $\cos \alpha = \pm \dfrac{1}{\sqrt{3}}$

=> Thus the direction cosines are: $\pm \dfrac{1}{\sqrt{3}}$, $\pm \dfrac{1}{\sqrt{3}}$, $\pm \dfrac{1}{\sqrt{3}}$.

问题10：如果是单位向量$\vec{a}$成一个角度$\dfrac{\pi}{3}$和$\hat{i}$ ， $\dfrac{\pi}{4}$和$\hat{j}$和一个锐角$\theta$和$\hat{k}$ ，然后找到\ theta，从而找到$\vec{a}$ 。

解决方案：

The unit vector be,

=> $\vec{r} = r_1\hat{i}+r_2\hat{j}+r_3\hat{k}$

=> $r_1 , r_2, r_3 = \cos (\dfrac{\pi}{3}), \cos (\dfrac{\pi}{4}), \cos \theta$

Given that $\vec{r}$ is unit vector,

=> $|\vec{r}| = 1$

=> $\sqrt{r_1^2+r_2^2+ r_3^2 }= 1$

=>$r_1^2+r_2^2+ r_3^2 = 1$

=>  $(\cos (\dfrac{\pi}{3}))^2 + (\cos (\dfrac{\pi}{4}))^2 + (\cos \theta)^2 =1$

=> $(\cos \theta)^2 = 1-[\dfrac{1}{2}+\dfrac{1}{4}]$

=> $(\cos \theta)^2 = 1- \dfrac{3}{4}$

=> $(\cos \theta)^2 = \dfrac{1}{4}$

=> $\cos \theta = \dfrac{1}{2}$

=> $\theta = \cos ^{-1} (\dfrac{1}{2})$

=> $\theta = \dfrac{\pi}{3}$

问题11：找到向量$\vec{r}$数量级$3\sqrt{2}$构成一个角度的单位$\dfrac{\pi}{4}$和$\dfrac{\pi}{2}$分别使用y和z轴。

解决方案：

Let l, m, n be the direction cosines of the vector $\vec{r}$.

We know that,

=> l² + m² + n² = 1

=> $\cos ^2 \alpha+ \cos ^2 \beta + \cos ^2 \gamma =1$

=> $l^2 =1 - [(\dfrac{1}{\sqrt{2}})^2+(0)^2]$

=> $l^2 = 1- \dfrac{1}{2}$

=> $l = \pm \dfrac{1}{\sqrt{2}}$

Thus vector is,

=> $\vec{r} = |\vec{r}|(l\hat{i}+m\hat{j}+n\hat{k})$

=> $\vec{r} = 3\sqrt{2}(\pm \dfrac{1}{\sqrt{2}} \hat{i} + \dfrac{1}{\sqrt{2}}\hat{j} + 0\hat{k})$

=> $\vec{r} = \pm 3\hat{i} + 3\hat{j}$

问题12：向量$\vec{r}$相对于3个轴倾斜的角度相等。如果大小$\vec{r}$是$2\sqrt{3}$ ， 找$\vec{r}$ 。

解决方案：

Let l, m, n be the direction cosines of the vector $\vec{r}$.

Given that the vector is inclined at equal angles to the 3 axes.

=> $l = m = n = \cos \alpha = \cos \beta = \cos \gamma$

We know that,

=> l² + m² + n² = 1

=> $3\cos ^2\alpha = 1$

=>$\cos \alpha = \pm \dfrac{1}{\sqrt{3}}$

Hence, the vector is given as,

=> $\vec{r} = |\vec{r}|(l\hat{i}+m\hat{j}+n\hat{k})$

=> $\vec{r} = 2\sqrt{3}(\pm \dfrac{1}{\sqrt{3}}\hat{i}+ \pm \dfrac{1}{\sqrt{3}}\hat{j}+ \pm \dfrac{1}{\sqrt{3}}\hat{k})$

=> $\vec{r} = \pm 2(\hat{i}+\hat{j}+\hat{k})$