第 12 类 RD Sharma 解决方案 – 第 23 章向量代数 – 练习 23.5

问题1.如果一个点(-4,-3)的位置向量是$\overrightarrow{a}$ ， 找$|\overrightarrow{a}|$ .

解决方案：

We have, 

$\overrightarrow{a} = -4\hat i - 3\hat j \\|\overrightarrow{a}| = \sqrt[]{(-4)^2+ (-3)^2} \\ = \sqrt[]{16+9} \\ = \sqrt[]{25} \\= 5 \\|\overrightarrow{a}| = 5$

问题2.如果位置向量$\overrightarrow{a}$点 (12,n) 是这样的$|\overrightarrow{a}|= 13$ ，找到值。

解决方案：

We have, 

$\overrightarrow{a} = 12\hat i + n \hat j \\ |\overrightarrow{a}| = \sqrt[]{12^2 + n^2} \\ 13 = \sqrt[]{144 + n^2}$

On squaring both sides, 

$(13)^2 = (\sqrt[]{144 + n^2})^2 \\169 = 144 + n^2 \\n^2 = 169 - 144 \\n^2 = 25 \\n^2 = \pm \sqrt[]{25} \\n = \pm 5$

问题 3. 找到一个与向量平行的大小为 4 个单位的向量$\sqrt[]{3} \hat i + \hat{j}$ .

解决方案：

Given, 

$\overline{a} = \sqrt{3 \hat i} + \hat{j}$

Let $\overline{b}$ is a vector parallel to $\overline{a}$
Therefore,
$\overline{b} = \lambda{\overline{a}}$ for any scalar
$\\ = \lambda(\sqrt{3} \hat i + \lambda \hat j) \\ = \overline{b} = \lambda \sqrt{3} \hat i + \lambda \hat j \\ |\overline{b}| = \sqrt{(\lambda \sqrt{3})^2+(\lambda)^2} \\ = \sqrt{4 \lambda^2} \\ |\overline{b}| = 2\lambda \\ 4 = 2\lambda \\ \lambda = 2 \\ \overline{b}= \lambda \sqrt{3} \hat i + \lambda \hat j \\ \overline{b}= 2 \sqrt{3} \hat i + \lambda \hat j$

问题 4. 快递$\overlinearrow{AB}$就单位向量而言 (i)A = (4,-1),B = (1,3) (ii)A = (-6,3) , B = (-2,-5)

解决方案：

(i) We have,
A = (4,-1)
B = (1,3)
Position Vector of A = $4\hat i - \hat j$
Position Vector of B = $\hat i + 3 \hat j$
Now,
$\overlinearrow {AB} = Position Vector of B - Position Vector of A \\ ( \hat i + 3 \hat j - 4\hat i + \hat j) \\ \overlinearrow{AB} = -3\hat i + 4 \hat j \\ |\overlinearrow{AB}| = \sqrt{(-3)^2 + (4)^2} \\ = \sqrt{9+16} \\ = \sqrt{25} \\ |\overlinearrow{AB}| = 5$
Therefore,
$\overlinearrow{AB} = -3\hat i + 4 \hat j$

(ii) We have,
A = (-6,3)
B = (-2,-5)
Position Vector of A = $-6\hat i + 3\hat j$
Position Vector of B = $-2\hat i - 5\hat j$
Now,
$\overlinearrow {AB} = Position Vector of B - Position Vector of A \\ (-2\hat i -5 \hat j) - (-6\hat i + 3\hat j) \\ \overlinearrow{AB} = 4\hat i - 8\hat j \\ |\overlinearrow{AB}| = \sqrt{(4)^2 + (-8)^2} \\ = \sqrt{16+64} \\ = \sqrt{80} \\ = \sqrt{16*5} \\ |\overlinearrow{AB}| = 4 \sqrt{5}$
Therefore,
$\overlinearrow{AB} = 4\hat i - 8\hat j$

问题 5. 求位置向量的尖端坐标，等价于$\overrightarrow{AB}$ , 其中 A 和 B 的坐标是 (-1,3) 和 (-2,1)

解决方案：

We have, 

A = (-1,3)

B = (-2,1)

Now, 

Position Vector of $A = -\hat i + 3 \hat j$

Position Vector of $-2 \hat i + 1\hat j$

Therefore, 

$\overrightarrow{AB} = Position Vector of B - Position Vector of A \\ = (-2\hat i + \hat j) - (-\hat i + 3 \hat j) \\ = -2 \hat i + \hat j + \hat i -3 \hat j \\ = -\hat i - 2 \hat j$

Coordinate of the position vector $\overrightarrow{AB} = -\hat i - 2 \hat j$

问题 6. ABCD 是平行四边形。如果A,B,C的坐标分别为(-2,-1),(3,0),(1,-2)，求D的坐标。

解决方案：

Here, A = (-2,-1)

B = (3,0)

C = (1,-2)

Let us assume D be (x , y).

Computing Position Vector of AB, we have,

= Position Vector of B – Position Vector of A

$= (3 \hat i) - (-2 \hat i - \hat j) \\ \overrightarrow{AB} = 5 \hat i + \hat j \\ \overrightarrow{DC} = Position Vector of C - Position Vector of D \\ = ( \hat i - 2 \hat j) - (x \hat i + y \hat j) \\ = \hat i - 2 \hat j - x \hat i - y \hat j \\ \overrightarrow{DC} = (1-x)\hat i + (-2-y)\hat j$

Comparing LHS and RHS of both, 

5 = 1-x 

x = -4 

And, 

1 = -2-y

y = -3

So, coordinates of D = (-4,-3).

问题 7. 如果点 A(3,4), B(5,-6) 和 C(4,-1) 的位置向量是$\vec{a}, \vec{b}, \vec{c}$分别计算$\vec{a} + 2\vec{b} - 3\vec{c}$ .

解决方案：

Computing the position vectors of all the points we have, 

$\vec{a} = 3\hat{i} + 4\hat{j} \\ \vec{b} = 5\hat{i} -6\hat{j} \\ \vec{c} = 4\hat{i} - \hat{j}$

Now, 

Computing the final value after substituting the values, 

$\vec{a} + 2\vec{b}-3\vec{c} = (3\hat i + 4\hat j ) + 2(5 \hat i - 6 \hat j) - 3(4 \hat i - \hat j) \\ = 3 \hat i + 4 \hat j + 10 \hat i - 12 \hat j -12 \hat i + 3 \hat j \\ = \hat i - 5 \hat j \\Therefore, \\ \vec{a} + 2\vec{b}-3\vec{c} = \hat i - 5 \hat j$

问题 8. 如果$\vec {a}$是其尖端为 (-5,3) 的位置向量，找到点 B 的坐标，使得$\overrightarrow{AB} = \vec {a}$ , A 的坐标为 (-4,1)。

解决方案：

Given,
Coordinate of A = (4,-1)
Position vector of A = $4\hat i - \hat j$
Position vector of $\vec{a} = 5\hat i - 3\hat j$
Let coordinate of point B = (x, y)
Position vector of B = $x\hat i + y\hat j$
Given that, $\overrightarrow{AB} = \vec{a}$
Position vector of B – Position vector of A = \vec{a}
$(x\hat i + y\hat j) - (4\hat i - \hat j) = 5\hat i - 3\hat j \\ (x - 4) \hat i + (y+1)\hat j = 5 \hat i - 3 \hat j$
Comparing the coefficients of LHS and RHS
x – y = 5
x = 9
Also,
y + 1 = 3
y = -1
So, coordinate of B = (9,-4)

问题 9. 证明点$2 \hat i , -\hat i -4 \hat j and -\hat i+4 \hat j$形成一个等腰三角形。

解决方案：

$|\overrightarrow{AB}| = 5 units \\ |\overrightarrow{BC}| = \sqrt[]{8^2} \\ |\overrightarrow{BC}| = 8 units \\ |\overrightarrow{AC}| = \sqrt[]{(-3)^2 + (8)^2} \\ = \sqrt[]{9+16} \\ = \sqrt[]{25} \\Here, \\|\overrightarrow{AB}| = |\overrightarrow{AC}|$

So, the two sides AB and AC of the triangle ABC are equal. 

Therefore, ABC is an isosceles triangle. 

问题 10. 求一个与向量平行的单位向量$\hat i + \sqrt{3} \hat j$ .

解决方案：

We have, 

Let $\vec {a} = \hat i + \sqrt{3} \hat j$

Suppose $\vec{a}$ is any vector parallel to $\vec{a}$

$\vec{b} = λ \vec{a}$ , where λ is any scalar.

$= λ(\hat i + \sqrt{3} \hat j)$

$\vec{b} = λ(\hat i + \sqrt{3} \hat j)$

Unit vector of

$\vec{b}= \frac{\vec{b}}{|\vec{b}|} \\ \hat{b} = \frac{\hat i + \sqrt[]{3} \hat j}{2} \\ \hat{b} = \frac{1}{2}(\hat i + \sqrt[]{3} \hat j)$
Therefore,
$\\ \hat{b} = \frac{1}{2}\hat i + \frac{\sqrt{3}}{2} \hat j$

问题 11. 找到以下每个点的位置向量沿坐标轴的分量：

(i) P(3,2)

(ii) Q(-5,1)

(iii) R(-11,-9)

(iv) S(4,-3)

解决方案：

(i) Given, P = (3,2)

Position vector of P = $3\hat i + 2\hat j$

Component of P along x-axis = $3 \hat i$

Component of P along y-axis = $2 \hat j$

(ii) Given, Q = (-5,1)

Position vector of Q = $-5\hat i + \hat j$

Component of Q along x-axis = $-5 \hat i$

Component of Q along y-axis =$\hat j$

(iii) Given, R = (-11,-9)

Position vector of R =$-11\hat i -9 \hat j$

Component of R along x-axis =$-11 \hat i$

Component of R along y-axis = $-9 \hat j$

(iv) Given, S = (4,-3)

Position vector of S =$4\hat i -3 \hat j$

Component of S along x-axis =$4 \hat i$

Component of S along y-axis = $-3 \hat j$