第12类RD Sharma解决方案–第23章，矢量代数–练习23.6 |套装1

问题1：求向量的大小$\vec{a} = 2\hat{i}+3\hat{j}-6\hat{k}$ 。

解决方案：

Magnitude of a vector $x\hat{i}+y\hat{j}+z\hat{k} = \sqrt{x^2+y^2+z^2}$

=>$|\vec{a}| = \sqrt{2^2+3^2+(-6)^2}$

=> $|\vec{a}| = \sqrt{4+9+36}$

=> $|\vec{a}| = \sqrt{49}$

=>$|\vec{a}| = 7$

问题2：在以下方向上找到单位向量$3\hat{i}+4\hat{j}-12\hat{k}$ 。

解决方案：

We know that unit vector of a vector $\vec{a}$ is given by,

=> $\hat{p} = \dfrac{\vec{a}}{|\vec{a}|}$

=> $\hat{p} = \dfrac{1}{\sqrt{3^2+4^2+(-12)^2}}(3\hat{i}+4\hat{j}-12\hat{k})$

=> $\hat{p} = \dfrac{1}{\sqrt{9+16+144}}(3\hat{i}+4\hat{j}-12\hat{k})$

=> $\hat{p} = \dfrac{1}{\sqrt{169}}(3\hat{i}+4\hat{j}-12\hat{k})$

=> $\hat{p} = \dfrac{1}{13}(3\hat{i}+4\hat{j}-12\hat{k})$

问题3：在向量结果的方向上找到一个单位向量$\hat{i}-\hat{j}+3\hat{k}$ ， $2\hat{i}+\hat{j}-2\hat{k}$和$\hat{i}+2\hat{j}-2\hat{k}$ 。

解决方案：

Let,

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

=> $\vec{b} = 2\hat{i}+\hat{j}-2\hat{k}$

=> $\vec{c} = \hat{i}+2\hat{j}-2\hat{k}$

Let $\vec{d}$ be the resultant,

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

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

=> $\vec{d} = 4\hat{i}+2\hat{j}-\hat{k}$

Unit vector is,

=> $\hat{p} = \dfrac{\vec{d}}{|\vec{d}|}$

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

=> $\hat{p} = \dfrac{1}{\sqrt{16+4+1}}(4\hat{i}+2\hat{j}-\hat{k})$

=> $\hat{p} = \dfrac{1}{\sqrt{21}}(4\hat{i}+2\hat{j}-\hat{k})$

问题4：平行四边形的相邻边由矢量表示$\vec{a}=\hat{i}+\hat{j}-\hat{k}$和$\vec{b}=-2\hat{i}+\hat{j}+2\hat{k}$ 。找到平行于平行四边形对角线的单位向量。

解决方案：

Let PQRS be the parallelogram.

Given that, PQ = $\hat{i}+\hat{j}-\hat{k}$ and QR = $-2\hat{i}+\hat{j}+2\hat{k}$.

Thus, the diagonals are: PR and SQ.

=> $\vec{PR} = \vec{PQ} + \vec{QR}$

=> $\vec{PR} = \vec{a} + \vec{b}$

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

=> $\vec{PR} = -\hat{i} + 2\hat{j} +\hat{k}$

=> $\vec{SQ} = \vec{PQ} - \vec{PS}$

=> $\vec{SQ} = \vec{a} - \vec{b}$

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

=> $\vec{SQ} = 3\hat{i}-3\hat{k}$

Thus the unit vectors in the direction of the diagonals are:

=> $\hat{PR} = \dfrac{\vec{PR}}{|\vec{PR}|}$

=> $\hat{PR} = \dfrac{1}{\sqrt{(-1)^2+2^2+1^2}}( -\hat{i} + 2\hat{j} +\hat{k})$

=> $\hat{PR} = \dfrac{1}{\sqrt{6}}(-\hat{i}+2\hat{j}+\hat{k})$

=> $\hat{SQ} = \dfrac{\vec{SQ}}{|\vec{SQ}|}$

=> $\hat{SQ} = \dfrac{1}{\sqrt{3^2+(-3)^2}}( 3\hat{i}-3\hat{k})$

=> $\hat{SQ} = \dfrac{1}{3\sqrt{2}}(3\hat{i}-3\hat{k})$

问题5：如果$\vec{a} = 3\hat{i}-\hat{j}-4\hat{k}$ ， $\vec{b}= -2\hat{i}+4\hat{j}-3\hat{k}$和$\vec{c}=\hat{i}+2\hat{j}-\hat{k}$ ， 找$|3\vec{a}-2\vec{b}+4\vec{c}|$ 。

解决方案：

Given, $\vec{a} = 3\hat{i}-\hat{j}-4\hat{k}$, $\vec{b}= -2\hat{i}+4\hat{j}-3\hat{k}$ and $\vec{c}=\hat{i}+2\hat{j}-\hat{k}$.

Let,

=> $\vec{d} = 3\vec{a}-2\vec{b}+4\vec{c}$

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

=> $\vec{d} = (9\hat{i}-3\hat{j}-12\hat{k})+ (4\hat{i}-8\hat{j}+6\hat{k})+(4\hat{i}+8\hat{j}-4\hat{k})$

=> $\vec{d} = 17\hat{i}-3\hat{j}-10\hat{k}$

The magnitude is given by,

=> $|\vec{d}| = \sqrt{17^2+(-3)^2+(-10)^2}$

=> $|\vec{d}| = \sqrt{289+9+100}$

=>$|\vec{d}| = \sqrt{398}$

问题6：如果$\vec{PQ} = 3\hat{i}+2\hat{j}-\hat{k}$并且P的坐标为(1，-1,2)，找到Q的坐标。

解决方案：

Given, $\vec{PQ} = 3\hat{i}+2\hat{j}-\hat{k}$

And, $\vec{P} = \hat{i}-\hat{j}+2\hat{k}$

=> $\vec{PQ} = \vec{Q}-\vec{P}$

=>$\vec{Q} = \vec{PQ}+ \vec{P}$

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

=> $\vec{Q} = 4\hat{i}+\hat{j}+\hat{k}$

=> Thus the coordinates of Q are (4,1,1).

问题7：证明要点$\hat{i}-\hat{j}$ ， $4\hat{i}-3\hat{j}+\hat{k}$和$2\hat{i}-4\hat{j}+5\hat{k}$是直角三角形的顶点。

解决方案：

Let,

=>$\vec{A} = \hat{i}-\hat{j}$

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

=> $\vec{C} = 2\hat{i}-4\hat{j}+5\hat{k}$

Thus, the 3 sides of the triangle are,

=> $\vec{AB} = \vec{B} - \vec{A}$

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

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

=> $\vec{BC} = \vec{C} - \vec{B}$

=> $\vec{BC} = (2\hat{i}-4\hat{j}+5\hat{k})-(4\hat{i}-3\hat{j}+\hat{k})$

=> $\vec{BC} = -2\hat{i}-\hat{j}+4\hat{k}$

=> $\vec{CA} = \vec{A} -\vec{C}$

=> $\vec{CA} = ( \hat{i}-\hat{j})-(2\hat{i}-4\hat{j}+5\hat{k})$

=> $\vec{CA} = -\hat{i}+3\hat{j}-5\hat{k}$

The lengths of every side are given by their magnitude,

=> $|\vec{AB}| = \sqrt{3^2+(-2)^2+1^2} = \sqrt{14}$

=>$|\vec{BC}| = \sqrt{(-2)^2+(-1)^2+4^2} = \sqrt{21}$

=>$|\vec{CA}| = \sqrt{(-1)^2+3^2+(-5)^2} = \sqrt{35}$

As we can see,

=>$|\vec{CA}|^2 = |\vec{AB}|^2+|\vec{BC}|^2$

=> These 3 points form a right-angled triangle.

问题8：如果三角形ABC的顶点A，B和C是具有位置矢量的点$a_1\hat{i}+a_2\hat{j}+a_3\hat{k}$ ， $b_1\hat{i}+b_2\hat{j}+b_3\hat{k}$ ， $c_1\hat{i}+c_2\hat{j}+c_3\hat{k}$分别由其边决定的向量是什么?找出这些向量的长度。

解决方案：

Let,

=> $\vec{a} =a_1\hat{i}+a_2\hat{j}+a_3\hat{k}$

=> $\vec{b} = b_1\hat{i}+b_2\hat{j}+b_3\hat{k}$

=> $\vec{c} = c_1\hat{i}+c_2\hat{j}+c_3\hat{k}$

The sides of the triangle are given as,

=> $\vec{AB} = \vec{b} - \vec{a}$

=> $\vec{AB} = ( b_1\hat{i}+b_2\hat{j}+b_3\hat{k})-(a_1\hat{i}+a_2\hat{j}+a_3\hat{k})$

=> $\vec{AB} = (b_1-a_1)\hat{i}+(b_2-a_2)\hat{j}+(b_3-a_3)\hat{k}$

=> $\vec{BC} = \vec{c}-\vec{b}$

=> $\vec{BC} = ( c_1\hat{i}+c_2\hat{j}+c_3\hat{k})-(b_1\hat{i}+b_2\hat{j}+b_3\hat{k})$

=> $\vec{BC} = (c_1-b_1)\hat{i}+(c_2-b_2)\hat{j}+(c_3-b_3)\hat{k}$

=> $\vec{CA} = \vec{a}-\vec{c}$

=> $\vec{CA} = ( a_1\hat{i}+a_2\hat{j}+a_3\hat{k})-(c_1\hat{i}+c_2\hat{j}+c_3\hat{k})$

=> $\vec{CA} = (a_1-c_1)\hat{i}+(a_2-c_2)\hat{j}+(a_3-c_3)\hat{k}$

The lengths of the sides are,

=>$|\vec{AB}| = \sqrt{(b_1-a_1)^2+(b_2-a_2)^2+(b_3-a_3)^2}$

=> $|\vec{BC}| = \sqrt{(c_1-b_1)^2+(c_2-b_2)^2+(c_3-b_3)^2}$

=> $|\vec{CA}| = \sqrt{(a_1-c_1)^2+(a_2-c_2)^2+(a_3-c_3)^2}$

问题9：找到从原点O到顶点为(1，-1,2)，(2,1,3)和(-1,2，-1)的三角形质心的向量。

解决方案：

The position of the centroid is given by,

=> (x, y, z) = $(\dfrac{x_1+x_2+x_3}{3},\dfrac{y_1+y_2+y_3}{3},\dfrac{z_1+z_2+z_3}{3})$

=> (x, y, z) = $(\dfrac{1+2+(-1)}{3},\dfrac{(-1)+1+2}{3},\dfrac{2+3+(-1)}{3})$

=> (x, y, z) = $(\dfrac{2}{3},\dfrac{2}{3},\dfrac{4}{3})$

The vector to the centroid from O is,

=> $\vec{c} = \dfrac{2}{3}\hat{i}+\dfrac{2}{3}\hat{j}+\dfrac{4}{3}\hat{k}$

问题10：找到点R的位置向量，该点将线段连接点p( $\hat{i}+2\hat{j}+\hat{k}$ )和q( $-\hat{i}+\hat{j}+\hat{k}$ )的比例为2：1。

(i)内部

解决方案：

The position vectors of a point that divides a line segment internally are given by,

=> $\vec{OR} = \dfrac{m\vec{Q}+n\vec{P}}{m+n}$, where $\dfrac{m}{n}=\dfrac{2}{1}$

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

=> $\vec{OR} = \dfrac{-\hat{i}+4\hat{j}+3\hat{k}}{3}$

(ii)从外部

解决方案：

The position vectors of a point that divides a line segment externally are given by,

=> $\vec{OR} = \dfrac{m\vec{Q}-n\vec{P}}{m-n}$, where $\dfrac{m}{n}=\dfrac{2}{1}$

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

=> $\vec{OR} = 3\hat{i}-\hat{k}$