第12类RD Sharma解决方案–第25章矢量或叉积–练习25.1 |套装3

问题25.如果$|\vec{a}|=\sqrt{26}$ ， $|\vec{b}|= 7$和$|\vec{a}\times\vec{b}|=35$ ， 找$\vec{a}.\vec{b}$

解决方案：

We know that,

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

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

=> $35 = \sqrt{26}\times7|\sin\theta|\times1$

=> $\sin\theta = \dfrac{35}{7\sqrt{26}}$

=> $\sin\theta = \dfrac{5}{\sqrt{26}}$

As $\cos^2\theta + \sin^2\theta =1$,

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

=> $\cos\theta = \sqrt{1-(\dfrac{5}{\sqrt{26}})^2}$

=>$\cos\theta = \sqrt{1-\dfrac{25}{26}}$

=> $\cos\theta = \dfrac{1}{\sqrt{26}}$

Thus,

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

=>$\vec{a}.\vec{b} = 7\sqrt{26}\times\dfrac{1}{\sqrt{26}}$

=> $\vec{a}.\vec{b} = 7$

问题26.找出由O，A，B形成的三角形的面积$\vec{OA} = \hat{i}+2\hat{j}+3\hat{k}$ ， $\vec{OB} = -3\hat{i}-2\hat{j}+\hat{k}$

解决方案：

The area of a triangle whose adjacent sides are given by $\vec{a}$ and $\vec{b}$ is$\dfrac{1}{2}|\vec{a}\times\vec{b}|$

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

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

=> $\vec{OA}\times\vec{OB} = \hat{i}[2+6]-\hat{j}[1+9]+\hat{k}[-2+6]$

=> $\vec{OA}\times\vec{OB} = 8\hat{i}-10\hat{j}+4\hat{k}$

=> Area = $\dfrac{1}{2} |\vec{OA}\times\vec{OB}|$

=> Area = $\dfrac{1}{2}\sqrt{8^2+(-10^2)+4^2}$

=> Area = $\dfrac{1}{2}\sqrt{64+100+16}$

=> Area = $\dfrac{1}{2}\times6\sqrt{45}$

=> Area = $3\sqrt{5}$ square units.

问题27.让$\vec{a}=\hat{i}+4\hat{j}+2\hat{k}$ ， $\vec{b}=3\hat{i}-2\hat{j}+7\hat{k}$和$\vec{c} = 2\hat{i}-\hat{j}+4\hat{k}$ 。找到一个向量$\vec{d}$两者垂直$\vec{a}$和$\vec{b}$和$\vec{c}.\vec{d} = 15$

解决方案：

Given that $\vec{d}$ is perpendicular to both $\vec{a}$ and $\vec{b}$.

=> $\vec{d}.\vec{a} =0$ ……….(1)

=> $\vec{d}.\vec{b} =0$ ……….(2)

Also,

=> $\vec{c}.\vec{d} = 15$ …….(3)

Let $\vec{d} = d_1\hat{i}+d_2\hat{j}+d_3\hat{k}$

From eq(1),

=> d₁ + 4d₂ + 2d₃ = 0 

From eq(2),

=> 3d₁ – 2d₂ + 7d₃ = 0

From eq(3),

=> 2d₁ – d₂ + 4d₃ = 15 

On solving the 3 equations we get,

d₁ = 160/3, d₂ = -5/3, and d₃ = -70/3, 

=> $\vec{d} = \dfrac {1}{3}(160\hat{i}-5\hat{j}-70\hat{k})$

问题28.找到一个垂直于每个向量的单位向量$\vec{a}+\vec{b}$和$\vec{a}-\vec{b}$ ， 在哪里$\vec{a}=3\hat{i}+2\hat{j}+2\hat{k}$和$\vec{b} = \hat{i}+2\hat{j}-2\hat{k}$ 。

解决方案：

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

Let $\vec{c} = \vec{a}+\vec{b}$

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

=> $\vec{c} = \hat{i}[3+1] +\hat{j}[2+2] +\hat{k}[2-2]$

=> $\vec{c} = 4\hat{i} +4\hat{j}$

Let $\vec{d} = \vec{a}-\vec{b}$

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

=> $\vec{d} = \hat{i}[3-1] +\hat{j}[2-2] +\hat{k}[2+2]$

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

A vector perpendicular to both $\vec{c}$ and $\vec{d}$ is,

=> $\vec{c}\times\vec{d} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\c_1 & c_2 & c_3\\d_1 & d_2 & d_3\end{vmatrix}$

=> $\vec{c}\times\vec{d} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\4 & 4 & 0\\2 & 0 & 4\end{vmatrix}$

=> $\vec{c}\times\vec{d} = \hat{i}[16-0]-\hat{j}[16-0]+\hat{k}[0-8]$

=> $\vec{c}\times\vec{d} = 16\hat{i}-16\hat{j}-8\hat{k}$

To find the unit vector,

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

=> $\hat{p} = \dfrac{1}{\sqrt{16^2+(-16)^2+(-8)^2}}(16\hat{i}-16\hat{j}-8\hat{k})$

=> $\hat{p} = \dfrac{1}{\sqrt{256+256+64}}(16\hat{i}-16\hat{j}-8\hat{k})$

=> $\hat{p} = \dfrac{1}{24}(16\hat{i}-16\hat{j}-8\hat{k})$

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

问题29.使用向量，找到顶点为A(2，3，5)，B(3，5，8)和C(2，7，8)的三角形的面积。

解决方案：

Given, A(2, 3, 5), B(3, 5, 8), and C(2, 7, 8)

Let,

=>$\vec{a} = A = 2\hat{i}+3\hat{j}+5\hat{k}$

=> $\vec{b} = B = 3\hat{i}+5\hat{j}+8\hat{k}$

=>$\vec{c} = C = 2\hat{2}+7\hat{j}+8\hat{k}$

Then,

=> $\vec{AB} = \vec{b}-\vec{a}$

=> $\vec{AB} = (3\hat{i}+5\hat{j}+8\hat{k})-(2\hat{i}+3\hat{j}+5\hat{k})$

=> $\vec{AB} = \hat{i}[3-2]+\hat{j}[5-3]+\hat{k}[8-5]$

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

=> $\vec{AC} = \vec{c}-\vec{a}$

=> $\vec{AC} = (2\hat{2}+7\hat{j}+8\hat{k})-(2\hat{i}+3\hat{j}+5\hat{k})$

=> $\vec{AC} = \hat{i}[2-2]+\hat{j}[7-3]+\hat{k}[8-5]$

=> $\vec{AC} = 4\hat{j}+3\hat{k}$

The area of a triangle whose adjacent sides are given by $\vec{a}$ and $\vec{b}$ is $\dfrac{1}{2}|\vec{a}\times\vec{b}|$

=> $\vec{AB}\times\vec{AC}= \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1 & 2& 3\\0 & 4 & 3\end{vmatrix}$

=> $\vec{AB}\times\vec{AC} = \hat{i}[6-12]-\hat{j}[3-0]+\hat{k}[4-0]$

=> $\vec{AB}\times\vec{AC} = -6\hat{i}-3\hat{j}+4\hat{k}$

=> Area = $\dfrac{1}{2}|\vec{AB}\times\vec{AC}|$

=> Area = $\dfrac{1}{2}\sqrt{(-6)^2+(-3)^2+4^2}$

=> Area = √61/2

问题30.如果$\vec{a}=2\hat{i}-3\hat{j}+\hat{k}$ ， $\vec{b}=-\hat{i}+\hat{k}$ ， $\vec{c}=2\hat{j}-\hat{k}$是三个向量，求出具有对角线的平行四边形的面积$(\vec{a}+\vec{b})$和$(\vec{b}+\vec{c})$ 。

解决方案：

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

Let,

=> $\vec{c} = (\vec{a}+\vec{b})$

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

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

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

=>$\vec{d} = (\vec{b}+\vec{c})$

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

=> $\vec{d} = -\hat{i}+2\hat{j}$

The area of the parallelogram having diagonals $\vec{c}$ and $\vec{d}$ is $\dfrac{1}{2}|\vec{c}\times\vec{d}|$

=> $\vec{c}\times\vec{d} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1 & -3& 2\\-1 & 2 & 0\end{vmatrix}$

=> $\vec{c}\times\vec{d} = \hat{i}[0-4] -\hat{j}[0+2]+\hat{k}[2-3]$

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

=> Area = $\dfrac{1}{2}|\vec{c}\times\vec{d}|$

=> Area = $\dfrac{1}{2}\sqrt{(-4)^2+(-2)^2+(-1)^2}$

=> Area = $\dfrac{1}{2}\sqrt{21}$

=> Area = √21/2

问题31.平行四边形的两个相邻边是$2\hat{i}-4\hat{j}+5\hat{k}$和$\hat{i}-2\hat{j}-3\hat{k}$ 。找到平行于其对角线之一的单位向量。另外，找到其区域。

解决方案：

Given a parallelogram ABCD and its 2 sides AB and BC.

By triangle law of addition,

=> $\vec{AC} = \vec{AB}+\vec{BC}$

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

=> $\vec{AC} = \hat{i}[2+1] +\hat{j}[-4-2]+\hat{k}[5-3]$

=> $\vec{AC} = 3\hat{i}-6\hat{j}+2\hat{k}$

Unit vector is,

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

=> $\hat{p} = \dfrac{1}{\sqrt{3^2+(-6)^2+2^2}}(3\hat{i}-6\hat{j}+2\hat{k})$

=> $\hat{p} = \dfrac{1}{\sqrt{49}}(3\hat{i}-6\hat{j}+2\hat{k})$

=> $\hat{p} = \dfrac{1}{7}(3\hat{i}-6\hat{j}+2\hat{k})$

Area of a parallelogram whose adjacent sides are given is $|\vec{a}\times\vec{b}|$

=> $|\vec{AB}\times\vec{BC}| = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2 & -4 & 5\\1 & -2 & -3\end{vmatrix}$

=> $|\vec{AB}\times\vec{BC}| = \hat{i}[12+10]-\hat{j}[-6-5]+\hat{k}[-4+4]$

=> $|\vec{AB}\times\vec{BC}| = 22\hat{i}+11\hat{j}$

Thus area is,

=> Area = $|22\hat{i}+11\hat{j}|$

=> Area = $\sqrt{22^2+11^2}$

=> Area = $\sqrt{605}$

=> Area = 11 √5 square units

问题32.如果有$\vec{a}=0$或者$\vec{b}=0$ ， 然后$\vec{a}\times\vec{b}=\vec{0}$ 。反过来是真的吗?举例说明。

解决方案：

Let us take two parallel non-zero vectors $\vec{a}$ and $\vec{b}$

=> $\vec{a}\times\vec{b} = \vec{0}$

For example,

$\vec{a} = \hat{i}$and $\vec{b}=2\hat{i}$

=> $\vec{a}\times\vec{b} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1 & 0 & 0\\2 & 0 & 0\end{vmatrix}$

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

But,

=> $|\vec{a}| = \sqrt{1^2} =1$

=> $|\vec{b}| = \sqrt{2^2} =2$

Hence the converse may not be true.

问题33.如果$\vec{a} = a_1\hat{i}+a_2\hat{j}+a_3\hat{k}$ ， $\vec{b} = b_1\hat{i}+b_2\hat{j}+b_3\hat{k}$和$\vec{c} = c_1\hat{i}+c_2\hat{j}+c_3\hat{k}$ ，然后确认$\vec{a}\times(\vec{b}\times\vec{c})=\vec{a}\times\vec{b}+\vec{a}\times\vec{c}$ 。

解决方案：

Given, $\vec{a} = a_1\hat{i}+a_2\hat{j}+a_3\hat{k}$, $\vec{b} = b_1\hat{i}+b_2\hat{j}+b_3\hat{k}$and $\vec{c} = c_1\hat{i}+c_2\hat{j}+c_3\hat{k}$

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

=> $\vec{a}\times(\vec{b}\times\vec{c}) = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\a_1 & a_2 & a_3\\(b_1+c_1) & (b_2+c_2) & (b_3+c_3)\end{vmatrix}$

=> $\vec{a}\times(\vec{b}\times\vec{c}) = \hat{i}[a_2(b_3+c_3)-a_3(b_2+c_2)]-\hat{j}[a_1(b_3+c_3)-a_3(b_1+c_1)]+\hat{k}[a_1(b_2+c_2)-a_2(b_1+c_1)]$

=> $\vec{a}\times(\vec{b}\times\vec{c}) = \hat{i}[a_2b_3+a_2c_3-a_3b_2-a_3c_2]+\hat{j}[-a_1b_3-a_1c_3+a_3b_1+a_3c_1]+\hat{k}[a_1b_2+a_1c_2-a_2b_1-a_2c_1]$ …..eq(1)

Now,

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

=> $\vec{a}\times\vec{b} = \hat{i}[a_2b_3-b_2a_3]-\hat{j}[a_1b_3-b_1a_3]+\hat{k}[a_1b_2-b_1a_2]$

And,

=> $\vec{a}\times\vec{c} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\a_1 & a_2 & a_3\\c_1 & c_2 & c_3\end{vmatrix}$

=>$\vec{a}\times\vec{c} = \hat{i}[a_2b_3-c_2a_3]-\hat{j}[a_1c_3-c_1a_3]+\hat{k}[a_1c_2-c_1a_2]$

Thus, 

=> $\vec{a}\times\vec{b}+\vec{a}\times\vec{c} = (\hat{i}[a_2b_3-b_2a_3]-\hat{j}[a_1b_3-b_1a_3]+\hat{k}[a_1b_2-b_1a_2]) + (\hat{i}[a_2b_3-c_2a_3]-\hat{j}[a_1c_3-c_1a_3]+\hat{k}[a_1c_2-c_1a_2])$

=> $\vec{a}\times\vec{b}+\vec{a}\times\vec{c} = \hat{i}[a_2b_3+a_2c_3-a_3b_2-a_3c_2]+\hat{j}[-a_1b_3-a_1c_3+a_3b_1+a_3c_1]+\hat{k}[a_1b_2+a_1c_2-a_2b_1-a_2c_1]$ …eq(2)

Thus eq(1) = eq(2)

Hence proved.

问题34(i)。使用向量找到顶点A(1，1，2)，B(2，3，5)和C(1，5，5)的三角形区域。

解决方案：

Given, A(1, 1, 2), B(2, 3, 5), and C(1, 5, 5)

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

=>$\vec{b} = B = 2\hat{i}+3\hat{j}+5\hat{k}$

=>$\vec{c} = C = \hat{i}+5\hat{j}+5\hat{k}$

Now 2 sides of the triangle are given by,

=> $\vec{AB} = \vec{b}-\vec{a}$

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

=> $\vec{AB} = \hat{i}[2-1] +\hat{j}[3-1]+\hat{j}[5-2]$

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

=> $\vec{AC} = \vec{c}-\vec{a}$

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

=> $\vec{AC} = \hat{i}[1-1] +\hat{j}[5-1]+\hat{j}[5-2]$

=> $\vec{AC} = 4\hat{j}+3\hat{k}$

Area of the triangle whose adjacent sides are given is $\dfrac{1}{2}|\vec{a}\times\vec{b}|$

=> $\vec{AB}\times\vec{AC} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1 & 2 & 3\\0 & 4 & 3\end{vmatrix}$

=> $\vec{AB}\times\vec{AC} = \hat{i}[6-12]-\hat{j}[3-0]+\hat{k}[4-0]$

=> $\vec{AB}\times\vec{AC} = -6\hat{i}-3\hat{j}+4\hat{k}$

Thus area of the triangle is,

=> Area = $\dfrac{1}{2}\sqrt{(-6)^2+(-3)^2+4^2}$

=> Area = $\dfrac{1}{2}\sqrt{36+9+16}$

=> Area = √61/2

问题34(ii)。使用向量找到顶点A(1、2、3)，B(2，-1、4)和C(4、5，-1)的三角形区域。

解决方案：

Given, A(1, 2, 3), B(2, -1, 4), and C(4, 5, -1)

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

=> $\vec{b} = B = 2\hat{i}-1\hat{j}+4\hat{k}$

=> $\vec{c} = C = 4\hat{i}+5\hat{j}-1\hat{k}$

Now 2 sides of the triangle are given by,

=>$\vec{AB} = \vec{b}-\vec{a}$

=>$\vec{AB} = (2\hat{i}-1\hat{j}+4\hat{k})-(4\hat{i}+5\hat{j}-1\hat{k})$

=> $\vec{AB} = \hat{i}[2-1] +\hat{j}[-1-2]+\hat{j}[4-3]$

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

=>$\vec{AC} = \vec{c}-\vec{a}$

=> $\vec{AC} = (4\hat{i}+5\hat{j}-1\hat{k})-(\hat{i}+2\hat{j}+3\hat{k})$

=> $\vec{AC} = \hat{i}[4-1] +\hat{j}[5-2]+\hat{j}[-1-3]$

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

Area of the triangle whose adjacent sides are given is $\dfrac{1}{2}|\vec{a}\times\vec{b}|$

=> $\vec{AB}\times\vec{AC} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1 & -3 & 1\\3 & 3 & -4\end{vmatrix}$

=> $\vec{AB}\times\vec{AC} = \hat{i}[12-3]-\hat{j}[-4-3]+\hat{k}[3+9]$

=> $\vec{AB}\times\vec{AC} = 9\hat{i}+7\hat{j}+12\hat{k}$

Thus area of the triangle is,

=> Area = $\dfrac{1}{2}\sqrt{(9)^2+(7)^2+12^2}$

=> Area = $\dfrac{1}{2}\sqrt{81+49+144}$

=> Area = √274/2

问题35.找到所有的量纲向量$10\sqrt{3}$垂直于$\hat{i}+2\hat{j}+\hat{k}$和$-\hat{i}+3\hat{j}+4\hat{k}$ 。

解决方案：

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

A vector perpendicular to both $\vec{a}$ and $\vec{b}$ is,

=> $\vec{a}\times\vec{b} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1 & 2 & 1\\-1 & 3 & 4\end{vmatrix}$

=> $\vec{a}\times\vec{b} = \hat{i}[8-3]-\hat{j}[4+1]+\hat{k}[3+2]$

=> $\vec{a}\times\vec{b} = 5\hat{i}-5\hat{j}+5\hat{k}$

Unit vector is,

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

=>$\hat{p} = \dfrac{1}{\sqrt{5^2+(-5)^2+5^2}}(5\hat{i}-5\hat{j}+5\hat{k})$

=> $\hat{p} = \dfrac{1}{5\sqrt{3}}(5\hat{i}-5\hat{j}+5\hat{k})$

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

Now vectors of magnitude $10\sqrt{3}$ are given by,

=> $10\sqrt{3}\hat{p} = \pm10\sqrt{3}\times \dfrac{1}{\sqrt{3}}(\hat{i}-\hat{j}+\hat{k})$

=> Required vectors, $\pm10(\hat{i}-\hat{j}+\hat{k})$

问题36.平行四边形的相邻边是$2\hat{i}-4\hat{j}-5\hat{k}$和$2\hat{i}+2\hat{j}+3\hat{k}$ 。找到与其对角线平行的2个单位向量。另外，找到其平行四边形的面积。

解决方案：

Given, $\vec{AB}=2\hat{i}-4\hat{j}-5\hat{k}$ and $\vec{BC} = 2\hat{i}+2\hat{j}+3\hat{k}$

=> $\vec{AC} = \vec{AB} + \vec{BC}$

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

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

Unit vector is,

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

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

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

Area is given by $|\vec{AB}\times\vec{BC}|$,