Exercise 27.1
Question 21
दिया है (Given) $\vec{a}=\frac{1}{7}(2 \hat{i}+3 \hat{j}+6 \hat{k}), \vec{b}=\frac{1}{7}(3 \hat{i}-6 \hat{j}+2 \hat{k})$ तथा (and) $\vec{c}=\frac{1}{7}(6 \hat{i}+2 \hat{j}-3 \hat{k})$.
दिखाएँ कि $\vec{a}, \vec{b}, \vec{c}$ परस्पर लम्ब इकाई सदिश हैं तथा $\vec{a} \times \vec{b}=\vec{c}$
Sol :
$\vec{a}=\frac{2}{7} \hat{\imath}+\frac{3}{7} \hat{j}+\frac{6}{7} \hat{k}$
$\vec{b}=\frac{3}{7} \hat{i}-\frac{6}{7} \hat{j}+\frac{2}{7} \hat{k}$
$\vec{c}=\frac{6}{7} \hat{\imath}+\frac{2}{7} \hat{j}-\frac{3}{7} \hat{k}$
$|\vec{a}|=\sqrt{\left(\frac{2}{7}\right)^{2}+\left(\frac{3}{7}\right)^{2}+\left(\frac{6}{7}\right)^{2}}=$
$=\sqrt{\frac{4}{49}+\frac{9}{49}+\frac{36}{49}}=\sqrt{1}$=1
$|\vec{b}|=\sqrt{\left(\frac{3}{7}\right)^{2}+\left(-\frac{6}{7}\right)^{2}+\left(\frac{2}{7}\right)^{2}}$
=1
$|\vec{c}|=\sqrt{\left(\frac{6}{7}\right)^{2}+\left(\frac{2}{7}\right)^{2}+\left(\frac{-3}{7}\right)^{2}}=1$
$\vec{a} \cdot \vec{b}=\frac{6}{49}-\frac{18}{49}+\frac{12}{49}=\frac{6-18+12}{49}=0$
$\therefore \vec{a} \perp \vec{b}$
$\vec{b} \cdot \vec{c}=\frac{18}{49}-\frac{12}{49}-\frac{6}{49}=0$
∴$\vec{b} \perp \vec{c}$
$\vec{c} \cdot \vec{0}=\frac{12}{49}+\frac{6}{49}-\frac{18}{49}=0$
∴$\vec{c} \perp \vec{a}$
∴$\vec{a}, \vec{b}$ तथा $\vec{c}$ परसपर लंब इकाई सदिश है।
अब,$\vec{a} \times \vec{b}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ \frac{2}{7} & \frac{3}{7} & \frac{6}{7} \\ \frac{3}{7} & \frac{-6}{7} & \frac{2}{7}\end{array}\right|$
$=\frac{1}{49}\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 6 \\ 3 & -6 & 2\end{array}\right|$
$=\frac{1}{49}\left[\hat{i}\left|\begin{array}{cc}3 & 6 \\ -6 & 2\end{array}\right|-\hat{\jmath}\left|\begin{array}{cc}2 & 6 \\ 3 & 2\end{array}\right|+\hat{k}\left|\begin{array}{cc}2 & 3 \\ 3 & -6\end{array}\right|\right]$
$=\frac{1}{49}[\hat{i}(6+36)-\hat{j}(4-18)+\hat{k}(-12-9]]$
$=\frac{1}{49}[42 \hat{i}+14 \hat{j}-21 \hat{k}]$○
$=\frac{1}{49} \times 7[6 \hat{\jmath}+2 \hat{j}-3 \hat{k}]$
$\vec{a} \times \vec{b}=\vec{c}$
Question 22
यदि (If) $\vec{a}=7 \hat{i}+3 \hat{j}-5 \hat{k}, \vec{b}=2 \hat{i}+5 \hat{j}-\hat{k}$ तथा (and)
$\vec{c}=-\hat{i}+2 \hat{j}+4 \hat{k}$ तो सत्यापित करें कि (Then verify that)
$\vec{a} \times(\vec{b}+\vec{c})=\vec{a} \times \vec{b}+\vec{a} \times \vec{c}$
Sol :
LHS
$\vec{a} \times(\vec{b}+\vec{c})=(7 \hat{i}+3 \hat{j}-5 \hat{k}) \times(\hat{i}+7 \hat{j}+3 \hat{k})$
$=\left|\begin{array}{ccc}\hat{\imath} & \hat{j} & \hat{k} \\ 7 & 3 & -5 \\ 1 & 7 & 3\end{array}\right|$
RHS
$\vec{a} \times \vec{b}+\vec{a} \times \vec{c}=\left|\begin{array}{ccc}\hat{\imath} & \hat{j} & \hat{k} \\ 7 & 3 & -5 \\ 2 & 5 & -1\end{array}\right|+\left|\begin{array}{ccc}\hat{i} & \hat{\jmath} & \hat{k} \\ 7 & 3 & -5 \\ -1 & 2 & 4\end{array}\right|$
$=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 7 & 3 & -5 \\ 2+(-1) & 5+2 & -1+4\end{array}\right|=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 7 & 3 & -5 \\ 1 & 7 & 3\end{array}\right|$
$\therefore \vec{a} \times(\vec{b}+\vec{c})=\vec{a} \times \vec{b}+\vec{a} \times \vec{c}$
Question 23
माना कि (Let) $\vec{a}=a_{1} \hat{i}+a_{2} \hat{j}+a_{3} \hat{k}, \vec{b}=b_{1} \hat{i}+b_{2} \hat{j}+b_{3} \hat{k}$ तथा (and) $\vec{c}=c_{1} \hat{i}+c_{2} \hat{j}+c_{3} \hat{k}$. तो दिखाएँ कि (Then show that) $\vec{a} \times(\vec{b}+\vec{c})=\vec{a} \times \vec{b}+\vec{a} \times \vec{c} .$
Sol :
LHS
$\vec{a} \times(\vec{b}+\vec{c})=\left(a_{1} \hat{i}+a_{2} \hat{\jmath}+a_{3} \hat{k}\right) \times\left[\left(b_{1}+c_{1}\right) \hat{i}+\left(b_{2}+c_{2}\right) \hat{j}+(b_3+c_3)\hat{k}\right]$
$=\left|\begin{array}{ccc}\hat{i} & \hat{\jmath} & \hat{k} \\ a_{1} & a_{2} & a_{3} \\ b_{1}+c_{1} & b_{2}+c_{2} & b_{3}+c_{3}\end{array}\right|$
$=\left|\begin{array}{lll}\hat{i} & \hat{j} & \hat{k} \\ a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3}\end{array}\right|+\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ a_{1} & a_{2} & a_{3} \\ c_{1} & c_{2} & c_{3}\end{array}\right|$
$=\vec{a} \times \vec{b}+\vec{a} \times \vec{c}$
Question 24
यदि (If) $\vec{a}=2 \hat{i}+5 \hat{j}-7 \hat{k}, \vec{b}=-3 \hat{i}+4 \hat{j}+\hat{k}$ तथा (and)
$\vec{c}=\hat{i}-2 \hat{j}-3 \hat{k}$, दिखाएँ कि (show that) $(\vec{a} \times \vec{b}) \times \vec{c}$ तथा (and)
$\vec{a} \times(\vec{b} \times \vec{c})$. समान नहीं हैं। (are not same)
Sol :
$\vec{a} \times \vec{b}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & 5 & -7 \\ -3 & 4 & 1\end{array}\right|=\hat{i}\left|\begin{array}{cc}5 & -7 \\ 4 & 1\end{array}\right|-\hat{j}\left|\begin{array}{cc}2 & -7 \\ -3 & 1\end{array}\right|+\hat{k}\left|\begin{array}{cc}2 &5 \\ -3 & 4\end{array}\right|$
$=\hat{i}(5+28)-\hat{j}(2-21)+\hat{k}(8+15)$
$=33 \hat{i}+19 \hat{j}+23 \hat{k}$
$\vec{b} \times \vec{c}=\left|\begin{array}{ccc}\hat{\imath} & \hat{j} & \hat{k} \\ -3 & 4 & 1 \\ 1 & -2 & -3\end{array}\right|$
$=\hat{i}\left|\begin{array}{cc}4 & 1 \\ -2 & -3\end{array}\right|-\hat{j}\left|\begin{array}{cc}-3 & 1 \\ 1 & -3\end{array}\right|+\hat{k}\left|\begin{array}{cc}-3 & 4 \\ 1 & -2\end{array}\right|$
$=\hat{i}(-12+2)-\hat{j}(9-1)+\hat{k}(6-4)$
$=-10 \hat{i}-8 \hat{j}+2 \hat{k}$
$(\vec{a} \times \vec{b}) \times \vec{c}=\left|\begin{array}{ccc}\hat{\imath} & \hat{j} & \hat{k} \\ 33 & 19 & 23 \\ 1 & -2 & -3\end{array}\right|$
$=\hat{i}\left|\begin{array}{cc}19 & 23 \\ -2 & -3\end{array}\right|-\hat{j}\left|\begin{array}{cc}33 & 23 \\ 1 & -3\end{array}\right|+\hat{k}\left|\begin{array}{cc}33 & 19 \\ 1 & -2\end{array}\right|$
$=\hat{i}(-57+46)-\hat{j}(-99-23)+\hat{k}(-66-19)$
$=-11 \hat{i}+122 \hat{\jmath}-85 \hat{k}$
$\vec{a} \times(\vec{b} \times \vec{c})=\left|\begin{array}{ccc}\hat{\hat{\imath}} & \hat{\jmath} & \hat{k} \\ 2 & 5 & -7 \\ -10 & -8 & 2\end{array}\right|=\hat{\imath}\left|\begin{array}{cc}5 & -7 \\ -8 & 2\end{array}\right|-\hat{\jmath}\left|\begin{array}{cc}2 & -7 \\ -10 & 2\end{array}\right|+\hat{k}\left|\begin{array}{cc}2 & 5 \\ -10 & -8\end{array}\right|$
$=\hat{i}(10-56)-\hat{\jmath}(4-70)+\hat{k}(-16+50)$
$=-46 \hat{i}+66\hat{\jmath}+34 \hat{k}$
$\therefore(\vec{a} \times \vec{b}) \times \vec{c} \neq \vec{a} \times(\vec{b} \times \vec{c})$
Question 25
यदि (If) $\vec{a}=2 \hat{i}+2 \hat{j}-\hat{k}$, तथा (and) $\vec{b}=3 \hat{i}-\hat{j}-\hat{k}$ तथा (and) $\vec{c}=\hat{i}+2 \hat{j}-3 \hat{k}$. तो सत्यापित करें कि (then verify that)
$\vec{a} \times(\vec{b} \times \vec{c})=(\vec{a} \cdot \vec{c}) \vec{b}-(\vec{a} \cdot \vec{b}) \vec{c}$.
Sol :
$\vec{a}=2 \hat{i}+2 \hat{j}-\hat{k}, \vec{b}=3 \hat{i}-\hat{j}-\hat{k}, \vec{c}=\hat{i}+2 \hat{\jmath}-3 \hat{k}$
$\vec{b} \times \vec{c}=\left|\begin{array}{rrr}\hat{i} & \hat{\jmath} & \hat{k} \\ 3 & -1 & -1 \\ 1 & 2 & -3\end{array}\right|$
$=\hat{i}\left|\begin{array}{rr}-1 & -1 \\ 2 & -3\end{array}\right|-\hat{j}\left|\begin{array}{ll}3 & -1 \\ 1 & -3\end{array}\right|+\hat{k}\left|\begin{array}{cc}3 & -1 \\ 1 & 2\end{array}\right|$
$=\hat{i}(3+2)-\hat{j}(-9+1)+\hat{k}(6+1)$
$=5 \hat{i}+8 \hat{j}+7 \hat{k}$
$\vec{a} \cdot \vec{c}$
=2+4+3=9,
$ \vec{a} \cdot \vec{b}$
=6-2+1=5
LHS
$\vec{a} \times\left(\vec{b} \times \vec{c}\right)=\left|\begin{array}{ccc}\hat{\imath} & \hat{j} & \hat{k} \\ 2 & 2 & -1 \\ 5 & 8 & 7\end{array}\right|=\hat{i}\left|\begin{array}{cc}2 & -1 \\ 8 & 7\end{array}\right|-\hat{j}\left|\begin{array}{cc}2 & -1 \\ 5 & 7\end{array}\right|+\hat{k}\left|\begin{array}{cc}2 & 2 \\ 5 & 8\end{array}\right|$
$=\hat{\imath}(14+8)-\hat{j}(14+5)+\hat{k}(16-10)$
$=22 \hat{i}-19 \hat{j}+6 \hat{k}$
RHS
$(\vec{a} \cdot \vec{c}) \vec{b}-(\vec{a} \cdot \vec{b}) \vec{c}=9 \cdot(3 \hat{i}-\hat{j}-\hat{k})-5(\hat{i}+2 \hat{j}-3 \hat{k})$
$=27 \hat{i}-9\hat{j}-9 \hat{k}-5 \hat{i}-10 \hat{j}+15 \hat{k}$○
$=22 \hat{i}-19 \hat{j}+6 \hat{k}$
$\therefore \vec{a} \times(\vec{b} \times \vec{c})=(\vec{a} \cdot \overrightarrow{c}) \vec{b}-(\vec{a} \cdot \vec{b}) \vec{c}$
Question 26
$\mathrm{P}(-\hat{i}+2 \hat{j}+6 \hat{k})$ से $\mathrm{A}(2 \hat{i}+3 \hat{j}-4 \hat{k})$ तथा $\mathrm{B}(8 \hat{i}+6 \hat{j}-8 \hat{k})$ को मिलाने वाली रेखा पर खींचे गये लम्ब की लम्बाई प्राप्त करें।
Sol :
माना PD⟂AB है।
$\begin{aligned} \overrightarrow{A B} &=(8 \hat{i}+6 \hat{j}-8 \hat{k})-(2 \hat{i}+3 \hat{j}-4 \hat{k}) \\ &=8 \hat{i}+6 \hat{j}-8 \hat{k}-2 \hat{i}-3 \hat{j}+4 \hat{k} \\ \overrightarrow{A B} &=6 \hat{i}+3 \hat{j}-4 \hat{k} \end{aligned}$
$\begin{aligned} \overrightarrow{A P} &=(-\hat{\imath}+2 \hat{j}+6 \hat{k})-(2 \hat{\jmath}+3 \hat{j}-4 \hat{k}) \\ &=-\hat{i}+2 \hat{j}+6 \hat{k}-2 \hat{i}-3 \hat{j}+4 \hat{k} \\ \overrightarrow{A P} &=-3 \hat{i}-\hat{j}+10 \hat{k} \end{aligned}$
$\overrightarrow{A B} \times \overrightarrow{A P}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 6 & 3 & -4 \\ -3 & -1 & 10\end{array}\right|$
$=\hat{i}\left|\begin{array}{cc}3 & -4 \\ -1 & 10\end{array}\right|-\hat{j}\left|\begin{array}{cc}6 & -4 \\ -3 & 10\end{array}\right|+\hat{k}\left|\begin{array}{cc}6 & 3 \\ -3 & -1\end{array}\right|$
$=\hat{i}(30-4)-\hat{j}(60-12)+\hat{k}(-6+9)$
$=26 \hat{i}-48 \hat{j}+3 \hat{k}$
$|\overrightarrow{A B} \times \overrightarrow{AP}|=\sqrt{26^{2}+(-48)^{2}+3^{2}}$
$=\sqrt{676+2304+9}$
$=\sqrt{2989}=\sqrt{7 \times 7 \times 61}=7\sqrt{61}$
$|\overrightarrow{A B}|=\sqrt{6^{2}+3^{2}+(-4)^{2}}$
$=\sqrt{36+3+16}=\sqrt{61}$
$P D=\frac{|\overrightarrow{AB} \times \overrightarrow{A P}|}{|\overrightarrow{AB}|}=\frac{7 \sqrt{61}}{\sqrt{61}}$=7
Question 27
माना कि $\vec{a}=(3,-1,0)$ तथा $\vec{b}=\left(\frac{1}{2}, \frac{3}{2}, 1\right) \cdot \vec{a} \times \vec{c}=4 \vec{b}$ तथा $\vec{a} \cdot \vec{c}=1$ को संतुष्ट करता हुआ सदिश $\vec{c}$ ज्ञात करें।
Sol :
माना $\vec{a}=3 \hat{\imath}-\hat{\jmath}$ , $\vec{b}=\frac{1}{2} \hat{i}+\frac{3}{2} \hat{j}+\hat{k}$
$\vec{z}=x \hat{i}+y \hat{j}+z \hat{k}$
$\vec{a} \times \vec{c}=4 \vec{b}$
$\left|\begin{array}{rrr}\hat{i} & \hat{j} & \hat{k} \\ 3 & -1 & 0 \\ x & y & z\end{array}\right|=4\left(\frac{1}{2} \hat{i}+\frac{3}{2} \hat{\jmath}+\hat{k}\right)$
$i\left|\begin{array}{cc}-1 & 0 \\ y & z\end{array}\right|-\hat{\jmath}\left|\begin{array}{ll}3 & 0 \\ x & z\end{array}\right|+\hat{k}\left|\begin{array}{cc}3 & -1 \\ x & y\end{array}\right|=2 \hat{i}+6 \hat{j}+4 \hat{k}$
$\hat{i}(-z-0)-\hat{\jmath}(3 z-0)+\hat{k}(3{y}+x)=2 \hat{i}+6 \hat{j}+4 \hat{k}$
$-z \hat{i}-3 z \hat{j}+(x+3 y) \hat{k}=2 \hat{i}+6 \hat{j}+4 \hat{k}$
-z=2⇒z=-2, x+3y=4...(i)
∵$\vec{a} \cdot \vec{c}=1$
$(3 \hat{i}-\hat{\jmath}) \cdot(x \hat{i}+y \hat{j}+z \hat{k})=1$
3x-y=1....(ii)×3
$\begin{aligned}x+3y&=4\\9x-3y&=3 \\ \hline 10x&=7\end{aligned}$
$x=\frac{7}{10} \Rightarrow y=\frac{11}{10}$
∴$\vec{c}=\frac{7}{10} \hat{i}+\frac{11}{10} \hat{j}+(-2) \hat{k}$
$=\frac{7 \hat{i}+10 \hat{j}-20 \hat{k}}{10}$
$=\frac{1}{10}(7 \hat{i}+10 \hat{\jmath}-20 \hat{k})$
Question 28
यदि $\vec{a}=(0,1,-1)$ तथा $\vec{c}=(1,1,1)$ दिए गए सदिश हैं तो एक सदिश $\vec{b}$ ज्ञात करें जो $\vec{a} \times \vec{b}+\vec{c}=0$ तथा $\vec{a} \cdot \vec{b}=3$ को संतुष्ट करता है।
Sol :
माना $\vec{b}=x \hat{i}+y \hat{\jmath}+z \hat{k}$
$\vec{a}=\hat{\jmath}-\hat{k}, \vec{c}=\hat{i}+\hat{j}+\hat{k}$
$\vec{a} \times \vec{b}+\vec{c}=\vec{0}$
$\left|\begin{array}{rrr}\hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & -1 \\ x & y & z\end{array}\right|+(\hat{i}+\hat{j}+\hat{k})=\overrightarrow{0}$
$\hat{i}\left|\begin{array}{cc}1 & -1 \\ y & z\end{array}\right|-\hat{\jmath}\left|\begin{array}{cc}0 & -1 \\ x & z\end{array}\right|+\hat{k}\left|\begin{array}{ll}0 & 1 \\ x & y\end{array}\right|+\hat{i}+\hat{j}+\hat{k}=\overrightarrow{0}$
$\hat{\imath}(z+y)-\hat{j}(0+x)+\hat{k}(0-x)+\hat{i}+\hat{j}+\hat{k}=\overrightarrow{6}$
$\hat{i}(z+y)-\hat{j}(0+x)+\hat{k}(0-x)+\hat{\imath}+\hat{j}+\hat{k}=\vec{0}$
$(1+y+z) \hat{i}+(1-x) \hat{j}+(1-x) \hat{k}=0 \hat{i}+0 \hat{j}+0 \cdot \hat{k}$
1+y+z=0 , 1-x=0
y+z=-1...(i) , x=1
$\vec{a} \cdot \vec{b}=3 \Rightarrow(\hat{j}-\hat{k}) \cdot(x \hat{i}+y \hat{j}+z \hat{k})=3$
y-z=3
समीकरण (i) तथा (ii) से,
$\begin{aligned}y+z&=-1 \\y-z&=3 \\ \hline2 y&=2\end{aligned}$
y=1 , z=-2
∴$\vec{b}=1 \cdot \hat{i}+1 \cdot \hat{j}+(-2) \hat{k} \Rightarrow \vec{b}=\hat{i}+\hat{\jmath}-2 \hat{k}$
Question 29
दिखाएँ कि $(\vec{a}-\vec{d}) \times(\vec{b}-\vec{c})+(\vec{b}-\vec{d}) \times(\vec{c}-\vec{a})+(\vec{c}-\vec{d}) \times(\vec{a}-\vec{b}), \vec{d}$ से स्वतंत्र है ।
Sol :
$(\vec{a}-\vec{d}) \times(\vec{b}-\vec{c})+(\vec{b}-\vec{d}) \times(\vec{c}-\vec{a})+(\vec{c}-\vec{d}) \times(\vec{a}-\vec{b})$
$= \vec{a} \times \vec{b}-\vec{a} \times \vec{c}-\vec{a} \times \vec{b}+\vec{a} \times \vec{c}+\vec{b} \times \vec{c}-\vec{b} \times \vec{a}-\vec{d} \times \vec{c}$ $+\vec{d} \times \vec{a}+\vec{c} \times \vec{a}-\vec{c} \times \vec{b}-\vec{d} \times \vec{a}+\vec{d} \times \vec{b}$
$=\vec{a} \times \vec{b}+\vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a}+\vec{c} \times \vec{a}$
$=2 \vec{a} \times \vec{b}+2 \vec{b} \times \vec{c}+2 \vec{c} \times \vec{a}$
$=2(\vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a}), \vec{d}$ से स्वतंत्र है।
Question 30
(i) सिद्ध करें कि (Prove that)
$(\vec{a}+3 \vec{b}) \times(\vec{a}+\vec{b})+(3 \vec{a}-5 \vec{b}) \times(\vec{a}-\vec{b})=\overline{0}$
Sol :
LHS
$(\vec{a}+3 \vec{b}) \times(\vec{a}+\vec{b})+(3 \vec{a}-5 \vec{b})\times(\vec{a}-\vec{b})$
$=\vec{a} \times \vec{a}+\vec{a} \times \vec{b}+3 \vec{b} \times \vec{a}+3 \vec{b} \times \vec{b}+3 \vec{a} \times \vec{a}$ $-3 \vec{a} \times \vec{b}-5 \vec{b} \times \vec{a}+5 \vec{b} \times \vec{b}$
$=\overrightarrow{0}-2 \vec{a} \times \vec{b}-2 \vec{b} \times \vec{a}+3 \times \vec{0}+3 \times \overrightarrow{0}+5 \times \overrightarrow{0}$
$=-2 \vec{a} \times \vec{b}+2 \vec{a} \times \vec{b}$
=0
(ii) सिद्ध करें कि (Prove that) $|(\vec{a}+\vec{b}) \times(\vec{a}-\vec{b})|=2 a b$ यदि (if) $\vec{a} \perp \vec{b}$.
Sol :
यदि $\vec{a} \perp \vec{b} \Rightarrow|\vec{a} \times \vec{b}|=a b \sin 90^{\circ}$
$\Rightarrow|\vec{a} \times \vec{b}|=a b$
LHS
$|(\vec{a}+\vec{b}) \times(\vec{a}-\vec{b})|=\mid \vec{a} \times \vec{a}-\vec{a} \times \vec{b}+\vec{b} \times \vec{a}-\vec{b} \times \vec{b}|$
$=|\vec{0}-\vec{a} \times \vec{b}-\vec{a} \times \vec{b}-\vec{0}|$
$=|-2 \vec{a} \times \vec{b}|$
$=2|\vec{a} \times \vec{b}|$
=2ab
No comments:
Post a Comment