Exercise 27.1
Question 11
दो सदिशों \vec{a} और \vec{b} के तल पर लम्ब इकाई सदिश निकालें, जहाँ
(i) \vec{a}=\hat{i}-\hat{j} तथा (and) \vec{b}=\hat{j}+\hat{k}
(ii) \vec{a}=4 \hat{i}-\hat{j}+3 \hat{k} तथा (and) \vec{b}=-2 \hat{i}+\hat{j}-2 \hat{k}
Sol :
(i)
\vec{a}=\hat{i}-\hat{j} , \vec{b}=\hat{j}+\hat{k}
\vec{a} \times \vec{b}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 0 \\ 0 & 1 & 1\end{array}\right|
=\hat{i}\left|\begin{array}{cc}-1 & 0 \\ 1 & 1\end{array}\right|-\hat{j}\left|\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right|+\hat{k}\left|\begin{array}{cc}1 & -1 \\ 0 & 1\end{array}\right|
=\hat{\imath}(-1-0)-\hat{j}(1-0)+\hat{k}(1+0)
\hat{a} \times \hat{b}=-\hat{\imath}-\hat{j}+\hat{k}
माना \vec{c}=\vec{a} \times \vec{b}
|\vec{c}|=|\vec{a} \times \vec{b}|=\sqrt{(-1)^{2}+(-1)^{2}+1^{2}}=\sqrt{3}
=-\frac{1}{\sqrt{3}} \hat{i}-\frac{1}{\sqrt{3}} \hat{\jmath}+\frac{1}{\sqrt{3}} \hat{k}
Question 12
निम्नलिखित सदिशों में से प्रत्येक पर लम्ब एक इकाई (मात्रक) सदिश निकालें। [Find unit vectors perpendicular to each of the following vectors.]
(i) 2 \hat{i}+3 \hat{j}-\hat{k}, \hat{i}+2 \hat{j}+3 \hat{k}
(ii) 2 \hat{i}-\hat{j}-\hat{k}, 2 \hat{i}-\hat{j}+3 \hat{k}
(iii) 4 \hat{i}-\hat{j}+3 \hat{k}, 2 \hat{i}+2 \hat{j}-\hat{k}
Sol :
(i) माना \vec{a}=2 \hat{\imath}+3 \hat{\jmath}-\hat{k}, \vec{b}=\hat{i}+2 \hat{\jmath}+3 \hat{k}
\vec{a} \times \vec{b}=\left|\begin{array}{ccc}\hat{\imath} & \hat{\jmath} & \hat{k} \\ 2 & 3 & -1 \\ 1 & 2 & 3\end{array}\right|
=\hat{i}\left|\begin{array}{cc}3 & -1 \\ 2 & 3\end{array}\right|-\hat{j}\left|\begin{array}{rr}2 & -1 \\ 1 & 3\end{array}\right|+\hat{k}\left|\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right|
=\hat{i}(9+2)-\hat{j}(6+1)+\hat{k}(4-3)
\vec{a} \times \vec{b}=11 \hat{i}-7 \hat{j}+\hat{k}
|\vec{a} \times \vec{b}|=\sqrt{11^{2}+(-7)^{2}+1^{2}}
=\sqrt{121+49+1}=\sqrt{171}
=\sqrt{19 \times 9 }=3\sqrt{19}
\vec{a} तथा \vec{b} पर लंम्ब इकाई सदिश =\frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}
=\frac{11 \hat{i}-7 \hat{j}+\hat{k}}{3 \sqrt{19}}
=\frac{7}{3 \sqrt{19}}(11 \hat{\imath}-7 \hat{\jmath}+\hat{k})
Question 13
निम्नलिखित सदिशों में से प्रत्येक पर लम्ब एक सदिश निकालें।
[Find a vector which is perpendicular to each of the vectors in the following :]
(i) \hat{i}-\hat{j}+\hat{k} तथा (and) 2 \hat{i}+3 \hat{j}-\hat{k}
(ii) \hat{i}+\hat{j}-2 \hat{k} तथा (and) 2 \hat{i}-2 \hat{j}+\hat{k}
Sol :
(i) माना \vec{a}=\hat{\imath}-\hat{\jmath}+\hat{k}, \vec{b}=2 \hat{i}+3 \hat{\jmath}-\hat{k}
\vec{a} \times \vec{b}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 2 & 3 & -1\end{array}\right|
=\hat{i}\left|\begin{array}{cc}-1 & 1 \\ 3 & -1\end{array}\right|-\hat{j}\left|\begin{array}{cc}1 & 1 \\ 2 & -1\end{array}\right|+\hat{k}\left|\begin{array}{cc}1 & -1 \\ 2 & 3\end{array}\right|
=\hat{i}(1-3)-\hat{j}(-1-2)+\hat{k}(3+2)
=-2 \hat{\imath}+3 \hat{j}+5 \hat{k}
Question 14
सदिश (\vec{a}+\vec{b}) और (\vec{a}-\vec{b}) में से प्रत्येक के लंबवत् मात्रक सदिश ज्ञात कीजिए जहाँ \vec{a}=\bar{i}+\hat{j}+\hat{k}, \vec{b}=\hat{i}+2 \hat{j}+3 \hat{k} हैं।
Sol :
\vec{a}=\bar{i}+\hat{j}+\hat{k}, \vec{b}=\hat{i}+2 \hat{j}+3 \hat{k}
\vec{a}+\vec{b}=2 \hat{\imath}+3 \hat{\jmath}+4 \hat{k}
\vec{a}-\vec{b}=-\hat{j}-2 \hat{k}
(\vec{a}+\vec{b}) \times(\vec{a}-\vec{b})=\left|\begin{array}{ccc}\hat{\imath} & \hat{j} & k \\ 2 & 3 & 4 \\ 0 & -1 & -3\end{array}\right|
=\hat{i}\left|\begin{array}{cc}3 & 4 \\ -1 & -2\end{array}\right|-\hat{j}\left|\begin{array}{cc}2 & 4 \\ 0 & -2\end{array}\right|+\hat{k}\left|\begin{array}{cc}2 & 3 \\ 0 & -1\end{array}\right|
=\hat{i}(-6+4)-\hat{j}(-4-0)+\hat{k}(-2-0)
=-2 \hat{i}+4 \hat{j}-2 \hat{k}
|(\vec{a}+\vec{b}) \times(\vec{a}-\vec{b})|=\sqrt{(-2)^{2}+4^{2}+(-2)^{2}} =\sqrt{4+16+4}=\sqrt{24}
=2 \sqrt{6}
\vec{a}+\vec{b} तथा \vec{a}-\vec{b} पर लंब ईकाई सदिश=\frac{(\vec{a}+\vec{b})\times\left(\vec{a}-\vec{b}\right)}{[(\vec{a}+\vec{b}) \times(\vec{a}-\vec{b})]}
=\frac{-2 \hat{i}+4 \hat{j}-2 \hat{k}}{2 \sqrt{6}}
=\frac{-2}{2 \sqrt{6}} \hat{i}+\frac{4}{2 \sqrt{6}} \hat{j}-\frac{2}{2 \sqrt{6}} \hat{k}
=-\frac{1}{\sqrt{6}} \hat{i}+\frac{2}{\sqrt{6}} \hat{j}-\frac{1}{\sqrt{6}} \hat{k}
Question 15
सदिशों \hat{i}+2 \hat{j}+\hat{k} तथा 3 \hat{i}+\hat{j}-\hat{k} के बीच का कोण ज्ञात करें साथ ही दोनों सदिशों में से प्रत्येक पर लम्ब एक इकाई सदिश भी ज्ञात करें।
Sol :
माना \vec{a}=\hat{i}+2 \hat{j}+\hat{k}, \vec{b}=3 \hat{i}+\hat{j}-\hat{k}
\vec{a} \times \vec{b}=\left|\begin{array}{ccc}\hat{\imath} & \hat{J} & \hat{k} \\ 1 & 2 & 1 \\ 3 & 1 & -1\end{array}\right|
=\hat{\imath}\left|\begin{array}{cc}2 & 1 \\ 1 & -1\end{array}\right|-\hat{\jmath}\left|\begin{array}{cc}1 & 1 \\ 3 & -1\end{array}\right|+\hat{k}\left|\begin{array}{cc}1 & 2 \\ 3 & 1\end{array}\right|
=\hat{i}(-2-1)-\hat{j}(-1-3)+\hat{k}(1-6)
=-3 \hat{\imath}+4 \hat{\jmath}-5 \hat{k}
|\vec{a} \times \vec{b}|=\sqrt{(-3)^{2}+4^{2}+(-5)^{2}}=\sqrt{9+16+25}
=\sqrt{50}=5\sqrt{2}
\vec{a} तथा \vec{b} दोनो पर लंब इकाई सदिश=\frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}
=\frac{-3 \hat{\imath}+4 \hat{j}-5 \hat{k}}{5 \sqrt{2}}
=\frac{1}{5 \sqrt{2}}(-3 \hat{\imath}+4 \hat{\jmath}-5 \hat{k})
\vec{a}=\hat{i}+2 \hat{j}+\hat{k}, \vec{b}=3 \hat{i}+\hat{\jmath}-\hat{k}
|\vec{a}|= \sqrt{1^{2}+2^{2}+1^{2}}=\sqrt{6} , |\vec{b}|=\sqrt{3^{2}+1^{2}+(-1)^{2}}=\sqrt{9+1+1}=\sqrt{11}
\sin \theta=\frac{|\vec{a} \times \vec{b}|}{|\vec{a}||\vec{b}|}
\sin \theta=\frac{5 \sqrt{2}}{\sqrt{6}\times \sqrt{11}}
\theta=\sin ^{-1}\left(\frac{5}{\sqrt{33}}\right)
Question 16
Question 17
2 \hat{i}-\hat{j}+\hat{k} तथा 3 \hat{i}+4 \hat{j}-\hat{k} में से प्रत्येक पर लम्ब इकाई सदिश क्या है ? सिद्ध करें कि दोनों सदिशों के बीच के कोण का ज्या \sqrt{\frac{155}{156}} है।
Sol :
माना \vec{a}=2 \hat{\jmath}-\hat{\jmath}+\hat{k}, \vec{b}=3 \hat{\imath}+4 \hat{\jmath}-\hat{k}
\vec{a} \times \vec{b}=\left|\begin{array}{rrr}\hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 1 \\ 3 & 4 & -1\end{array}\right|
=\hat{i}\left|\begin{array}{cc}-1 & 1 \\ 4 & -1\end{array}\right|-\hat{j}\left|\begin{array}{rr}2 & 1 \\ 3 & -1\end{array}\right|+\hat{k}\left|\begin{array}{cc}2 & -1 \\ 3 & 4\end{array}\right|
=\hat{i}(1-4)-\hat{j}(-2-3)+\hat{k}(8+3)
=-3 \hat{i}+5 \hat{j}+11 \hat{k}
|\vec{a} \times \vec{b}|=\sqrt{(-3)^{2}+5^{2}+11^{2}}=\sqrt{9+25+121}=\sqrt{155}
|\vec{a}|=\sqrt{2^{2}+(-1)^{2}+1^{2}}=\sqrt{6},|\vec{b}|=\sqrt{3^{2}+4^{2}+(-1)^{2}}=\sqrt{9+16+1}=\sqrt{26}
\vec{a} तथा \vec{b} पर लंब इकाई सदिश
=\frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}=\frac{-3 \hat{\imath}+5 \hat{\jmath}+11 \hat{k}}{\sqrt{155}}
=\frac{1}{\sqrt{155}}(-3 \hat{i}+5 \hat{j}+11 \hat{k})
\sin \theta=\frac{|\vec{a} \times \vec{b}|}{|\vec{a}||\vec{b}|}
\sin \theta=\frac{\sqrt{155}}{\sqrt{6} \times \sqrt{26}}
\sin \theta=\frac{\sqrt{155}}{\sqrt{156}}
\sin \theta=\sqrt{\frac{155}{156}}
Question 18
यदि बिन्दुएँ A, B, C क्रमश: (1,0,-1),(0,1,-1) तथा (-1,0,1) हैं तो रेखाओं AB तथा AC के बीच का कोण ज्ञात करें।
Sol :
\overrightarrow{A B}=(0-1) \hat{i}+(1-0) \hat{\jmath}+(-1+1) \hat{k}
\overrightarrow{A B}=-\hat{i}+\hat{\jmath}
\overrightarrow{A C}=(-1-1) \hat{\imath}+(0-0) \hat{\jmath}+(1+1) \hat{k}
\overrightarrow{A C}=-2 \hat{\imath}+2 \hat{k}
\overrightarrow{A B} \times \overrightarrow{A C}=\left|\begin{array}{ccc}\hat{\imath} & \hat{j} & \hat{k} \\ -1 & 1 & 0 \\ -2 & 0 & 2\end{array}\right|
=\hat{i}\left|\begin{array}{ll}1 & 0 \\ 0 & 2\end{array}\right|-\hat{\jmath}\left|\begin{array}{cc}-1 & 0 \\ -2 & 2\end{array}\right|+\hat{k}\left|\begin{array}{cc}-1 & 1 \\ -2 & 0\end{array}\right|
=i(2-0)-\hat{\jmath}(-2+0)+\hat{k}(-0+2)
=2 \hat{\imath}+2 \hat{\jmath}+2 \hat{k}
|\overrightarrow{AB} \times \overrightarrow{A C}|=\sqrt{2^{2}+2^{2}+2^{2}}=\sqrt{12}=2 \sqrt{3}
|\overrightarrow{A B}|=\sqrt{(-1)^{2}+1^{2}}=\sqrt{2}
|AC|=\sqrt{(-2)^{2}+2^{2}}=\sqrt{8}
\sin \theta=\frac{|\overrightarrow{A B} \times \overrightarrow{A C}|}{|\overrightarrow{AB}||\overrightarrow{A C}|}
\sin \theta=\frac{2 \sqrt{3}}{\sqrt{2} \times \sqrt{8}}
\sin \theta=\frac{2 \sqrt{3}}{4}
\sin \theta=\frac{\sqrt{3}}{2}
Question 19
इकाई परिमाण वाले सदिश का घटक निकालें जो सदिशों 2 \hat{i}+\hat{j}-4 \hat{k} तथा 3 \hat{i}+\hat{j}-\hat{k} पर लम्ब है।
Sol :
माना \vec{a}=2 \hat{j}+\hat{j}-4 \hat{k}
\vec{b}=3 \hat{i}+\hat{j}-\hat{k}
\vec{a} तथा \vec{b} पर लंब सदिश =\vec{a} \times \vec{b}
=\left|\begin{array}{ccc}\hat{\imath} & \hat{j} & \hat{k} \\ 2 & 1 & -4 \\ 3 & 1 & -1\end{array}\right|
=\hat{i}\left|\begin{array}{cc}1 & -4 \\ 1 & -1\end{array}\right|-\hat{j}\left|\begin{array}{cc}2 & -4 \\ 3 & -1\end{array}\right|+\hat{k}\left|\begin{array}{cc}2 & 1 \\ 3 & 1\end{array}\right|
=\hat{i}(-1+4)-\hat{j}(-2+12)+\hat{k}(2-3)
=3 \hat{i}-10 \hat{j}-\hat{k}
|\vec{a} \times \vec{b}|=\sqrt{3^{2}+(-10)^{2}+(-1)^{2}}=\sqrt{9+100+1}=\sqrt{110}
\vec{a} तबा \vec{b} पर लंब इकाई सदिश =\frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}=\frac{3 \hat{i}-10 \hat{j}-\hat{k}}{\sqrt{110}}
=\frac{1}{\sqrt{110}}(3 \hat{i}-10 \hat{j}-\hat{k})
Question 20
यदि तीन बिन्दुओं \mathrm{A}, \mathrm{B}, \mathrm{C} के स्थिति सदिश क्रमश: 2 \hat{i}+4 \hat{j}-\hat{k}, \hat{i}+2 \hat{j}-3 \hat{k} तथा 3 \hat{i}+\hat{j}+2 \hat{k} हैं, तो तल \mathrm{ABC} पर लम्ब एक सदिश निकालें ।.
Sol :
\overrightarrow{B C}=(3 \hat{i}+\hat{j}+2 \hat{k})-\left(\hat{i}+2 \hat{j}-3 \hat{k}\right)
=3 \hat{i}+\hat{j}+2 \hat{k}-\hat{i}-2 \hat{\jmath}+3 \hat{k}
\overrightarrow{B C}=2 \hat{i}-\hat{j}+5 \hat{k}
\overrightarrow{B A}=(2 \hat{i}+4 \hat{j}-\hat{k})-(\hat{i}+2 \hat{j}-3 \hat{k})
=2 \hat{i}+4 \hat{j}-\hat{k}-\hat{\imath}-2 \hat{\jmath}+3 \hat{k}
\overrightarrow{B A}=\hat{\imath}+2 \hat{j}+2 \hat{k}
तल ABC पर लंब सदिश=\overrightarrow{B A} \times \overrightarrow{B C}
=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 2 \\ 2 & -1 & 5\end{array}\right|
=\hat{i}\left|\begin{array}{cc}2 & 2 \\ -1 & 5\end{array}\right|-\hat{j}\left|\begin{array}{ll}1& 2 \\ 2 & 5\end{array}\right|+\hat{k}\left|\begin{array}{cc}1 & 2 \\ 2 & -1\end{array}\right|
=\hat{i}(10+2)-\hat{j}(5-4)+\hat{k}(-1-4)☺
=12 \hat{i}-\hat{j}-5 \hat{k}
तल ABC पर लंब सदिश=-(12 \hat{i}-\hat{\jmath}-5 \hat{k})
=-12 \hat{\imath}+\hat{j}+5 \hat{k}
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