第12类RD Sharma解决方案–第25章矢量或叉积–练习25.1 |套装2

问题13 $|\vec{a}|=13$ ， $|\vec{b}|=5$和$\vec{a}.\vec{b}=60$ ， 找$|\vec{a}\times\vec{b}|$

解决方案：

We know that,

=> $\vec{a}.\vec{b} = |\vec{a}||\vec{b}|\cos\theta$

=>$|\vec{a}.\vec{b}| = |\vec{a}||\vec{b}||\sin\theta|$

=> $60 = 13\times5\times\cos\theta$

=> $65\cos\theta = 60$

=>$\cos\theta = \dfrac{12}{13}$

Also,

=>$| \vec{a}\times\vec{b}| = |\vec{a}||\vec{b}||\cos\theta||\hat{n}|$

And $\sin^2\theta + \cos^2\theta = 1$

=> $\sin\theta = \sqrt{1-\cos^2\theta}$

=> $\sin\theta = \sqrt{1-(\dfrac{12}{13})^2}$

=> $\sin\theta = \sqrt{\dfrac{25}{169}}$

=> $\sin\theta = \dfrac{5}{13}$

=> $|\vec{a}\times\vec{b}| = 13\times5\times\dfrac{5}{13}$

=>$| \vec{a}\times\vec{b}| = 25$

问题14：找到2个向量之间的角度$\vec{a}$和$\vec{b}$ ， 如果$|\vec{a}\times\vec{b}| = \vec{a}.\vec{b}$

解决方案：

Given $|\vec{a}\times\vec{b}| = \vec{a}.\vec{b}$

=>$|\vec{a}||\vec{b}|\sin\theta|\hat{n}| = |\vec{a}||\vec{b}|\cos\theta$

=>$|\vec{a}||\vec{b}|\sin\theta = |\vec{a}||\vec{b}|\cos\theta$, as $\hat{n}$ is a unit vector.

=> $\sin\theta = \cos\theta$

=> $\tan\theta = 1$

=> $\theta = \dfrac{\pi}{4}$

问题15 $\vec{a}\times\vec{b} = \vec{b}\times\vec{c} \neq \vec{0}$ ，然后证明$\vec{a}+\vec{c}=m\vec{b}$ ，其中m是任何标量。

解决方案：

Given that $\vec{a}\times\vec{b} = \vec{b}\times\vec{c} \neq \vec{0}$

=> $\vec{a}\times\vec{b} - \vec{b}\times\vec{c} = \vec{0}$

=> $\vec{a}\times\vec{b} -[-(\vec{c}\times\vec{a})] = \vec{0}$

=>$\vec{a}\times\vec{b} + \vec{c}\times\vec{b} = \vec{0}$

Using distributive property,

=> $(\vec{a}+\vec{c})\times\vec{b}=\vec{0}$

If two vectors are parallel, then their cross-product is 0 vector.

=> $(\vec{a}+\vec{c})$ and $\vec{b}$ are parallel vectors.

=> $(\vec{a}+\vec{c}) = m\vec{b}$

Hence proved.

问题16：如果$|\vec{a}|=2$ ， $|\vec{b}|=7$和$\vec{a}\times\vec{b} = 3\hat{i}+2\hat{j}+6\hat{k}$ ，找出两者之间的夹角$\vec{a}$和$\vec{b}$

解决方案：

Given that,$|\vec{a}|=2$, $|\vec{b}|=7$and $\vec{a}\times\vec{b} = 3\hat{i}+2\hat{j}+6\hat{k}$

We know that,

=> $\vec{a}\times\vec{b} =|\vec{a}||\vec{b}|\sin\theta\hat{n}$

=> $|\vec{a}\times\vec{b}| =|\vec{a}||\vec{b}|\sin\theta\|\hat{n}|$

=> $|\vec{a}\times\vec{b}| =|\vec{a}||\vec{b}|\sin\theta$

=> $\sqrt{3^2+2^2+6^2} = 2\times7\times\sin\theta$

=> $\sqrt{9+4+36} = 14\sin\theta$

=> $7 = 14\sin\theta$

=> $\sin\theta = \dfrac{1}{2}$

=> $\theta = \dfrac{\pi}{6}$

问题17：如果可以得出什么推论$\vec{a}\times\vec{b}=\vec{0}$和$\vec{a}.\vec{b}=0$

解决方案：

Given, $\vec{a}\times\vec{b}=\vec{0}$and $\vec{a}.\vec{b}=0$

=>$\vec{a}\times\vec{b} = \vec{0}$

=>$|\vec{a}|\vec{b}|\sin\theta\hat{n} = \vec{0}$

Either of the following conditions is true,

1.$|\vec{a}| = 0$

2. $|\vec{b}| =0$

3. $|\vec{a}| = |\vec{b}| = 0$

4.$\vec{a}$is parallel to $\vec{b}$

=> $\vec{a}.\vec{b}=0$

=> $|\vec{a}||\vec{b}|\cos\theta = 0$

Either of the following conditions is true,

1. $|\vec{a}| = 0$

2.$|\vec{b}| =0$

3. $|\vec{a}| = |\vec{b}| = 0$

4. $\vec{a}$is perpendicular to $\vec{b}$

Since both these conditions are true, that implies atleast one of the following conditions is true,

1. $|\vec{a}| = 0$

2. $|\vec{b}| =0$

3. $|\vec{a}| = |\vec{b}| = 0$

问题18 $\vec{a}$ ， $\vec{b}$和$\vec{c}$是3个单位向量$\vec{a}\times\vec{b} = \vec{c}$ ， $\vec{b}\times\vec{c}=\vec{a}$和$\vec{c}\times\vec{a} = \vec{b}$ 。显示$\vec{a}$ ， $\vec{b}$和$\vec{c}$形成单位向量的正交右手三合会。

解决方案：

Given, $\vec{a}\times\vec{b} = \vec{c}$, $\vec{b}\times\vec{c}=\vec{a}$and $\vec{c}\times\vec{a} = \vec{b}$

As,

=>$\vec{c} = \vec{a}\times\vec{b}$

=> $\vec{c}$ is perpendicular to both $\vec{a}$and $\vec{b}$.

Similarly,

=> $\vec{a}$ is perpendicular to both $\vec{b}$and $\vec{c}$

=> $\vec{b}$ is perpendicular to both $\vec{a}$and $\vec{c}$

=> $\vec{a}$, $\vec{b}$and $\vec{c}$ are mutually perpendicular.

As, $\vec{a}$, $\vec{b}$and $\vec{c}$ are also unit vectors,

=> $\vec{a}$, $\vec{b}$and $\vec{c}$ form an orthogonal right-handed triad of unit vectors

Hence proved.

问题19.找到一个垂直于平面ABC的单位矢量，其中A，B和C的坐标为A(3，-1，2)，B(1，-1，3)和C(4，- 3，1)。

解决方案：

Given A(3, -1, 2), B(1, -1, 3) and C(4, -3, 1).

Let,

=> $\vec{a} = A = 3\hat{i}- \hat{j} +2\hat{k}$

=> $\vec{b} = B = \hat{i} -\hat{j} + 3\hat{k}$

=> $\vec{c} = C = 4\hat{i}-3\hat{j}+\hat{k}$

Plane ABC has two vectors $\vec{AB}$ and $\vec{AC}$

=> $\vec{AB} = \vec{b} - \vec{a}$

=> $\vec{AB} = (\hat{i}-\hat{j}-3\hat{k})-(3\hat{i}-\hat{j}+2\hat{k})$

=> $\vec{AB}= (1-3)\hat{i}+ (-1+1)\hat{j} +(-3-2)\hat{k}$

=> $\vec{AB} = -2\hat{i}-5\hat{k}$

=> $\vec{AC} = \vec{c} - \vec{a}$

=> $\vec{AC} = (4\hat{i}-3\hat{j}+\hat{k})-(3\hat{i}-\hat{j}+2\hat{k})$

=> $\vec{AC} = (4-3)\hat{i}+ (-3+1)\hat{j} +(1-2)\hat{k}$

=> $\vec{AC} = \hat{i}-2\hat{i}-\hat{k}$

A vector perpendicular to both $\vec{AB}$and $\vec{AC}$is given by,

=> $\vec{a}\times\vec{b} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\a_1 & a_2 & a_3\\b_1 & b_ 2& b_3\end{vmatrix}$

=> $\vec{AB}\times\vec{AC}= \dfrac{1}{7}\times\dfrac{1}{7}\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\-2 & 0 & -5\\1 & -2& -1\end{vmatrix}$

=> $\vec{AB}\times\vec{AC} = \hat{i}[(0)(-1)-(-2)(-5)] -\hat{j}[(-2)(-1)-(1)(-5)] +\hat{k}[(-2)(-2)-(1)(0)]$

=> $\vec{AB}\times\vec{AC} = \hat{i}[0-10]-\hat{j}[2+5]+\hat{k}[4-0]$

=>$\vec{AB}\times\vec{AC} = -10\hat{j}-7\hat{j}+4\hat{k}$

To find the unit vector,

=> $\hat{p} = \dfrac{\vec{AB}\times\vec{AC}}{|\vec{AB}\times\vec{AC}|}$

=> $\hat{p} = \dfrac{1}{\sqrt{(-10^2)+(-7)^2+4^2}}(-10\hat{j}-7\hat{j}+4\hat{k})$

=> $\hat{p} = \dfrac{1}{\sqrt{100+49+16}}(-10\hat{j}-7\hat{j}+4\hat{k})$

=> $\hat{p} = \dfrac{1}{\sqrt{165}}(-10\hat{j}-7\hat{j}+4\hat{k})$

问题20.如果a，b和c是三角形ABC的边BC，CA和AB的长度，则证明$\vec{BC} +\vec{CA} +\vec{AB} = \vec{0}$并推断出$\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}$

解决方案：

Given that $|\vec{BC}|=a$, $|\vec{CA}|=b$and $|\vec{AB}| = c$

From triangle law of vector addition, we have

=> $\vec{AB} + \vec{BC} = \vec{AC}$

=> $\vec{AB} + \vec{BC} = -\vec{CA}$

=> $\vec{AB} + \vec{BC} + \vec{CA} = \vec{0}$

=> $\vec{BC} +\vec{CA} +\vec{AB} = \vec{0}$

=> $\vec{a} + \vec{b} + \vec{c} = \vec{0}$

=> $\vec{a}\times(\vec{a}+\vec{b}+\vec{c}) = \vec{a}\times\vec{0}$

=> $\vec{a}\times\vec{a} + \vec{a}\times\vec{b} + \vec{a}\times\vec{c} = \vec{0}$

=> $\vec{0} + \vec{a}\times\vec{b} + \vec{a}\times\vec{c} = \vec{0}$

=> $\vec{a}\times\vec{b} + \vec{a}\times\vec{c} = \vec{0}$

=> $\vec{a}\times\vec{b} = -\vec{a}\times\vec{c}$

=> $\vec{a}\times\vec{b} = \vec{c}\times\vec{a}$

=> $|\vec{a}||\vec{b}|\sin C = |\vec{c}||\vec{a}|\sin B$

=>$|\vec{b}|\sin C = |\vec{c}|\sin B$

=> $b\sin C = c\sin B$

=> $\dfrac{b}{\sin B} = \dfrac{c}{\sin C}$

Similarly,

=> $\vec{b}\times(\vec{a}+\vec{b}+\vec{c}) = \vec{b}\times\vec{0}$

=> $\vec{b}\times\vec{a} + \vec{b}\times\vec{b} + \vec{b}\times\vec{c} = \vec{0}$

=> $\vec{0} + \vec{b}\times\vec{a} + \vec{b}\times\vec{c} = \vec{0}$

=> $\vec{b}\times\vec{a} + \vec{b}\times\vec{c} = \vec{0}$

=> $\vec{b}\times\vec{a} = -\vec{b}\times\vec{c}$

=> $\vec{b}\times\vec{a} = \vec{c}\times\vec{b}$

=>$|\vec{b}||\vec{a}|\sin C = |\vec{c}||\vec{b}|\sin A$

=> $|\vec{a}|\sin C = |\vec{c}|\sin A$

=> $a\sin C = c\sin A$

=> $\dfrac{a}{\sin A} = \dfrac{c}{\sin C}$

=> $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{sin C}$

Hence proved.

问题21.如果$\vec{a} = \hat{i}-2\hat{j}+3\hat{k}$和$\vec{b}=2\hat{i}+3\hat{j}-5\hat{k}$ ，然后找到$\vec{a}\times\vec{b}$ 。验证$\vec{a}$和$\vec{a}\times\vec{b}$彼此垂直。

解决方案：

Given, $\vec{a} = \hat{i}-2\hat{j}+3\hat{k}$and $\vec{b}=2\hat{i}+3\hat{j}-5\hat{k}$

=>$\vec{a}\times\vec{b} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\a_1 & a_2 & a_3\\b_1 & b_2 & b_3\end{vmatrix}$

=>$\vec{a}\times\vec{b} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1 & -2 & 3\\2 & 3 & -5\end{vmatrix}$

=> $\vec{a}\times\vec{b} = \hat{i}[(-2)(-5)-(3)(3)]-\hat{j}[(1)(-5)-(2)(3)]+\hat{k}[(1)(3)-(2)(-2)]$

=>$\vec{a}\times\vec{b} = \hat{i}[10-9]-\hat{j}[-5-6]+\hat{k}[3+4]$

=> $\vec{a}\times\vec{b} = \hat{i}+11\hat{j}+7\hat{k}$

Two vectors are perpendicular if their dot product is zero.

=>$(\vec{a}\times\vec{b}).\vec{a} = (\hat{i}-2\hat{j}+3\hat{k}).(\hat{i}+11\hat{j}+7\hat{k})$

=> $(\vec{a}\times\vec{b}).\vec{a} = \hat{i}.\hat{i}-2\hat{j}.11\hat{j}+3\hat{k}.7\hat{k}$

=> $(\vec{a}\times\vec{b}).\vec{a} = 1-22+21$

=> $(\vec{a}\times\vec{b}).\vec{a} =0$

Hence proved.

问题22：如果$\vec{p}$和$\vec{q}$是形成一个角度的单位向量$30\degree$ ，找到具有以下特征的平行四边形的面积： $\vec{a}=\vec{p}+2\vec{q}$和$\vec{b}=2\vec{p}+\vec{q}$作为其对角线。

解决方案：

Given $\vec{p}$and $\vec{q}$forming an angle of $30\degree$.

Area of a parallelogram having diagonals $\vec{a}$ and $\vec{b}$ is $\dfrac{1}{2}|\vec{a}\times\vec{b}|$

=> $\vec{p}\times\vec{q} = |\vec{p}||\vec{q}|\sin 30\degree \hat{n}$

=> $\vec{p}\times\vec{q} = 1\times1\times\dfrac{1}{2}\times \hat{n}$

=>$\vec{p}\times\vec{q} = \dfrac{1}{2} \hat{n}$

Thus area is,

=> Area = $\dfrac{1}{2}|(\vec{p}+2\vec{q})\times( 2\vec{p}+\vec{q})|$

=> Area = $\dfrac{1}{2}|\vec{p}\times( 2\vec{p}+\vec{q})+2\vec{q}\times( 2\vec{p}+\vec{q})|$

=> Area = $\dfrac{1}{2}|\vec{p}\times\vec{q}+4(\vec{q}\times\vec{p})|$

=> Area = $\dfrac{1}{2}|\vec{p}\times\vec{q}+4(-\vec{p}\times\vec{q})|$

=> Area = $\dfrac{1}{2}|-3(\vec{p}\times\vec{q})|$

=> Area =$\dfrac{3}{2}|(\vec{p}\times\vec{q})|$

=> Area = $\dfrac{3}{2}|\dfrac{1}{2} \hat{n}|$

=> Area = $\dfrac{3}{2}\times\dfrac{3}{2}\times1$

=> Area = $\dfrac{3}{4}$ square units

问题23：对于任何两个向量$\vec{a}$和$\vec{b}$ ， 证明$|\vec{a}\times\vec{b}|^2 = \begin{vmatrix} \vec{a}.\vec{a}& \vec{a}.\vec{b}\\\vec{b}.\vec{a} & \vec{b}.\vec{b}\end{vmatrix}$

解决方案：

We know that,

=> $\vec{a}\times\vec{b}= |\vec{a}||\vec{b}|\sin\theta\hat{n}$

=>$|\vec{a}\times\vec{b}|= |\vec{a}||\vec{b}|\sin\theta|\hat{n}|$

=>$|\vec{a}\times\vec{b}|= |\vec{a}||\vec{b}|\sin\theta$

=>$|\vec{a}\times\vec{b}|^2= |\vec{a}|^2|\vec{b}|^2\sin^2\theta$

=> $|\vec{a}\times\vec{b}|^2= |\vec{a}|^2|\vec{b}|^2(1-\cos^2\theta)$

=> $|\vec{a}\times\vec{b}|^2= |\vec{a}|^2|\vec{b}|^2 -(|\vec{a}|^2|\vec{b}|^2\cos^2\theta)$

=> $|\vec{a}\times\vec{b}|^2= |\vec{a}|^2|\vec{b}|^2 -(|\vec{a}|.|\vec{b}|)^2$

=>$|\vec{a}\times\vec{b}|^2= (\vec{a}.\vec{a})(\vec{b}.\vec{b})-(\vec{a}.\vec{b})(\vec{b}.\vec{a})$

=>$|\vec{a}\times\vec{b}|^2 = \begin{vmatrix} \vec{a}.\vec{a}& \vec{a}.\vec{b}\\\vec{b}.\vec{a} & \vec{b}.\vec{b}\end{vmatrix}$

Hence proved.

问题24.定义$\vec{a}\times\vec{b}$并证明$|\vec{a}\times\vec{b}|=(\vec{a}.\vec{b})\tan \theta$ ， 在哪里$\theta$是之间的角度$\vec{a}$和$\vec{b}$

解决方案：

Definition of $\vec{a}\times\vec{b}$: Let $\vec{a}$ and $\vec{b}$be 2 non-zero, non-parallel vectors. Then $\vec{a}\times\vec{b}$, is defined as a vector with the magnitude of $|\vec{a}||\vec{b}|\sin\theta$, and which is perpendicular to both the vectors $\vec{a}$ and $\vec{b}$.

We know that,

=> $\vec{a}\times\vec{b} = |\vec{a}||\vec{b}|\sin\theta\hat{n}$

=> $|\vec{a}\times\vec{b}| = |\vec{a}||\vec{b}|\sin\theta|\hat{n}|$

=>$|\vec{a}\times\vec{b}| = |\vec{a}||\vec{b}|\sin\theta$ ……………..(eq.1)

And as,

=> $\vec{a}.\vec{b} = |\vec{a}||\vec{b}|\cos\theta$

=>$|\vec{a}||\vec{b}| = \dfrac{\vec{a}.\vec{b}}{\cos\theta}$

Substituting in (eq.1),

=> $|\vec{a}\times\vec{b}| = \dfrac{\vec{a}.\vec{b}}{\cos\theta} \sin\theta$

=> $|\vec{a}\times\vec{b}| = (\vec{a}.\vec{b})\tan\theta$