第12类RD Sharma解决方案–第23章，矢量代数–练习23.6 |套装2

问题11：找到连接点P( $2\hat{i}-3\hat{j}+4\hat{k}$ )和Q( $4\hat{i}+\hat{j}-2\hat{k}$ )。

解决方案：

The mid-point of the line segment joining 2 vectors is given by:

=> $\vec{R} = \dfrac{\vec{P}+\vec{Q}}{2}$

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

=> $\vec{R} = \dfrac{6\hat{i}-2\hat{j}+2\hat{k}}{2}$

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

问题12：在向量的方向上找到单位向量$\vec{PQ}$ ，其中P和Q是点(1,2,3)和(4,5,6)。

解决方案：

Let,

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

=>$\vec{q} = 4\hat{i} + 5\hat{j}+6\hat{k}$

=> $\vec{PQ} = \vec{q}-\vec{p}$

=> $\vec{PQ} = (4\hat{i} + 5\hat{j}+6\hat{k})-(\hat{i}+2\hat{j}+3\hat{k})$

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

Unit vector is,

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

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

=> $\hat{PQ} = \dfrac{1}{3\sqrt{3}}3(\hat{i}+\hat{j}+\hat{k})$

=> $\hat{PQ} = \dfrac{1}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})$

问题13：证明点A( $2\hat{i}-\hat{j}+\hat{k}$ )，B( $\hat{i}-3\hat{j}-5\hat{k}$ )， C( $3\hat{i}-4\hat{j}-4\hat{k}$ )是直角三角形的顶点。

解决方案：

Let,

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

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

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

The line segments are,

=> $\vec{AB} = \vec{b}-\vec{a}$

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

=> $\vec{AB} = -\hat{i}-2\hat{j}-6\hat{k}$

=> $\vec{BC} = \vec{c}-\vec{b}$

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

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

=> $\vec{CA} = \vec{a}-\vec{c}$

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

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

The magnitudes of the sides are,

=> $|\vec{AB}| = \sqrt{(-1)^2+(-2)^2+(-6)^2} = \sqrt{41}$

=> $|\vec{BC}| = \sqrt{2^2+(-1)^2+1^2} = \sqrt{6}$

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

As we can see that $|\vec{AB}|^2 = |\vec {BC}|^2+|\vec{CA}|^2$

=> Thus, ABC is a right-angled triangle.

问题14：找到连接点P(2，3，4)和Q(4，1，-2)的向量中点的位置向量。

解决方案：

Let,

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

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

 The mid-point of the line segment joining 2 vectors is given by:

=> $\vec{r} = \dfrac{\vec{p}+\vec{q}}{2}$

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

=> $\vec{r} = \dfrac{6\hat{i}+4\hat{j}+2\hat{k}}{2}$

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

问题15：求x的值，其中x( $\hat{i}+\hat{j}+\hat{k}$ )是单位向量。

解决方案：

The magnitude of the given vector is,

=> $|x(\hat{i}+\hat{j}+\hat{k})| = \sqrt{x^2+x^2+x^2}$

=> $|x(\hat{i}+\hat{j}+\hat{k})| = \sqrt{3x^2}$

=> $|x(\hat{i}+\hat{j}+\hat{k})| = \pm x\sqrt{3}$

For it to be a unit vector,

=> $|x(\hat{i}+\hat{j}+\hat{k})| = 1$

=> $x\sqrt{3} = \pm 1$

=> $x = \pm \dfrac{1}{\sqrt{3}}$

问题16：如果$\vec{a}=\hat{i}+\hat{j}+\hat{k}$ ， $\vec{b} = 2\hat{i}-\hat{j}+3\hat{k}$和$\vec{c}=\hat{i}-2\hat{j}+\hat{k}$ ，找到与之平行的单位向量$2\vec{a}-\vec{b}+3\vec{c}$ 。

解决方案：

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

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

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

Thus, the unit vector is,

=> $\hat{p} = \dfrac{2\vec{a}-\vec{b}+3\vec{c}}{|2\vec{a}-\vec{b}+3\vec{c}|}$

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

=> $\hat{p} = \dfrac{1}{\sqrt{22}}(3\hat{i}-3\hat{j}+2\hat{k})$

问题17：如果$\vec{a}=\hat{i}+\hat{j}+\hat{k}$ ， $\vec{b}=4\hat{i}-2\hat{j}+3\hat{k}$和$\vec{c} = \vec{i}-2\hat{j}+\hat{k}$ ，找到一个大小为6个单位的向量，该向量与该向量平行$2\vec{a}-\vec{b}+3\vec{c}$ 。

解决方案：

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

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

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

Unit vector in that direction is,

=> $\hat{p} = \dfrac{2\vec{a}-\vec{b}+3\vec{c}}{|2\vec{a}-\vec{b}+3\vec{c}|}$

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

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

Given that the vector has a magnitude of 6,

=> Required vectors are : $\pm6\times\dfrac{1}{3}(\hat{i}-2\hat{j}+2\hat{k})$ = $\pm2(\hat{i}-2\hat{j}+2\hat{k})$

问题18：找到一个与向量的结果平行的幅度为5个单位的向量$\vec{a}=2\hat{i}+3\hat{j}-\hat{k}$和$\vec{b} = \hat{i}-2\hat{j}+\hat{k}$ 。

解决方案：

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

The resultant vector will be given by,

=> $\vec{r} = \vec{a}+\vec{b}$

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

=> $\vec{r} = 3\hat{i}+\hat{j}$

Unit vector is,

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

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

=> $\hat{p} = \dfrac{1}{\sqrt{10}}(3\hat{i}+\hat{j})$

Given that the vector has a magnitude of 5,

=> Required vectors are: $\pm5\times\dfrac{1}{\sqrt{10}}(3\hat{i}+\hat{j})= \pm\dfrac{5}{\sqrt{10}}(3\hat{i}+\hat{j})$

问题19：两个向量$\hat{j}+\hat{i}$和$3\hat{i}+\hat{j}+4\hat{k}$代表双方$\vec{AB}$和$\vec{AC}$分别为三角形ABC。通过A查找中位数的长度。

解决方案：

Let D be the point on BC, on which the median through A touches.

D is also the mid-point of BC.

The median $\vec{AD}$ is thus given by:

=> $\vec{AD} = \dfrac{\vec{B}+\vec{C}}{2}- \vec{A}$

=> $\vec{AD} = \dfrac{\vec{B}-\vec{A}+\vec{C}-\vec{A}}{2}$

=>$\vec{AD} = \dfrac{\vec{AB}+\vec{AC}}{2}$

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

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

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

Thus, the length of the median is,

=> $|\vec{AD}| = \sqrt{2^2+1^2+2^2}$

=> $|\vec{AD}| = \sqrt{9}$

=> $|\vec{AD}| = 3$ units