These NCERT Solutions for Class 12 Maths Chapter 10 सदिश बीजगणित Ex 10.4 Questions and Answers are prepared by our highly skilled subject experts.
NCERT Solutions for Class 12 Maths Chapter 10 सदिश बीजगणित Ex 10.4
प्रश्न 1.
यदि \(\vec{a}\) = \(\hat{i}\) – 7 \(\hat{j}\) + 7 \(\hat{k}\) और \(\vec{b}\) = 3 \(\hat{i}\) – 2 \(\hat{j}\) + 2 \(\hat{k}\), तो \(|\vec{a}\) × \(\vec{b}|\) ज्ञात कीजिए।
हल:
प्रश्नानुसार,
\(\vec{a}\) = \(\hat{i}\) – 7 \(\hat{j}\) + 7 \(\hat{k}\)
\(\vec{b}\) = 3 \(\hat{i}\) – 2 \(\hat{j}\) + 2 \(\hat{k}\)
तब
\(\vec{a}\) × \(\vec{b}\) = \(\left|\begin{array}{rrr}
\hat{i} & \hat{j} & \hat{k} \\
1 & -7 & 7 \\
3 & -2 & 2
\end{array}\right|\)
= \(\left|\begin{array}{ll}
-7 & 7 \\
-2 & 2
\end{array}\right| \hat{i}-\left|\begin{array}{ll}
1 & 7 \\
3 & 2
\end{array}\right| \hat{j}+\left|\begin{array}{ll}
1 & -7 \\
3 & -2
\end{array}\right| \hat{k}\)
= [-14 – (-2) × 7] \(\hat{i}\) – [1 × 2 – 3 × 7] \(\hat{j}\)
+ [1 × (-2) – 3(-7)] \(\hat{k}\)
= (-14 + 14) \(\hat{i}\) – (2 – 21) \(\hat{j}\) + (-2 + 21) \(\hat{k}\)
= 0 \(\hat{i}\) + 19 \(\hat{j}\) + 19 \(\hat{k}\) = 19 \(\hat{j}\) + 19 \(\hat{k}\)
अब
\(|\vec{a} \times \vec{b}|\) = \(|19 \hat{j}+19 \hat{k}|\)
= \(\sqrt{19^2+19^2}\)
= \(\sqrt{361+361}\) = \(\sqrt{722}\) = 19 \(\sqrt{2}\)
प्रश्न 2.
\(\vec{a}\) + \(\vec{b}\) और \(\vec{a}\) – \(\vec{b}\) की लम्ब दिशा में मात्रक सदिश ज्ञात कीजिए, जहाँ
\(\vec{a}\) = 3 \(\hat{i}\) + 2 \(\hat{j}\) + 2 \(\hat{k}\) और \(\vec{b}\) = \(\hat{i}\) + 2 \(\hat{j}\) – 2 \(\hat{k}\) हैं।
हल:
प्रश्नानुसार,
\(\vec{a}\) = 3 \(\hat{i}\) + 2 \(\hat{j}\) + 2 \(\hat{k}\)
तथा \(\vec{b}\) = \(\hat{i}\) + 2 \(\hat{j}\) – 2 \(\hat{k}\)
तब \(\vec{a}\) + \(\vec{b}\) = (3 \(\hat{i}\) + 2 \(\hat{j}\) + 2 \(\hat{k}\)) + (\(\hat{i}\) + 2 \(\hat{j}\) – 2 \(\hat{k}\))
= (3 + 1) \(\hat{i}\) + (2 + 2) \(\hat{j}\) + (2 – 2) \(\hat{k}\)
या
\(\vec{a}\) + \(\vec{b}\) = 4 \(\hat{i}\) + 4 \(\hat{j}\)
तथा
\(\vec{a}\) – \(\vec{b}\) = (3 \(\hat{i}\) + 2 \(\hat{j}\) + 2 \(\hat{k}\))-(\(\hat{i}\) + 2 \(\hat{j}\) – 2 \(\hat{k}\))
= (3 – 1) \(\hat{i}\) + (2 – 2) \(\hat{j}\) + (2 + 2) \(\hat{k}\)
या
\(\vec{a}\) – \( \vec{b}\) = 2 \(\hat{i}\) + 4 \(\hat{k}\)
अब (\(\vec{a}\) + \(\vec{b}\)) × (\(\vec{a}\) – \(\vec{b}\)) = \(\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 4 & 4 & 0 \\ 2 & 0 & 4\end{array}\right|\)
= \(\left|\begin{array}{ll}4 & 0 \\ 0 & 4\end{array}\right| \hat{i}-\left|\begin{array}{ll}4 & 0 \\ 2 & 4\end{array}\right| \hat{j}+\left|\begin{array}{ll}4 & 4 \\ 2 & 0\end{array}\right| \hat{k}\)
= 16 \(\hat{i}\) – 16 \(\hat{j}\) – 8 \(\hat{k}\)
∴ (\(\vec{a}\) – \(\vec{b}\)) तथा (\(\vec{a}\) – \(\vec{b}\)) की लम्ब दिशा में मात्रक सदिश
प्रश्न 3.
यदि एक मात्रक सदिश \(\vec{a}\), \(\hat{i}\) के साथ \(\frac{\pi}{3}\), \(\hat{j}\) के साथ \(\frac{\pi}{4}\) और \(\hat{k}\) के साथ एक न्यूनकोण θ बनाता है, तो \(\vec{a}\) का मान ज्ञात कीजिए और इसकी सहायता से \(\vec{a}\) के घटक भी ज्ञात कीजिए।
हल:
प्रश्न 4.
दर्शाइए कि:
(\(\vec{a}\) – \(\vec{b}\)) × (\(\vec{a}\) + \(\vec{b}\)) = 2(\(\vec{a}\) × \(\vec{b}\))
हल:
बायाँ पक्ष (L.H.S.)
= (\(\vec{a}\) – \(\vec{b}\)) × (\(\vec{a}\) + \(\vec{b}\))
= \(\vec{a}\) × (\(\vec{a}\) + \(\vec{b}\)) – \(\vec{b}\) × (\(\vec{a}\) + \(\vec{b}\))
= \(\vec{a}\) × \(\vec{a}\) + \(\vec{a}\) × \(\vec{b}\) – \(\vec{b}\) × \(\vec{a}\) + \(\vec{b}\) × \(\vec{b}\)
= \(\overrightarrow{0}\) + \(\vec{a}\) × \(\vec{b}\) – {-(\(\vec{a}\) × \(\vec{b}\))} + \(\overrightarrow{0}\)
= \(\vec{a}\) × \(\vec{b}\) + \(\vec{a}\) × \(\vec{b}\)
[∴ \(\vec{a}\) × \(\vec{b}\) = -(\(\vec{b}\) × \(\vec{a}\))
\(\vec{a}\) × \(\vec{a}\) = \(\overrightarrow{0}\)
\(\vec{b}\) × \(\vec{b}\) = \(\overrightarrow{0}\)]
= 2(\(\vec{a}\) × \(\vec{b}\)) = दायाँ पक्ष (R.H.S.)
प्रश्न 5.
λ और µ ज्ञात कीजिए, यदि
(2 \(\hat{i}\) + 6 \(\hat{j}\) + 27 \(\hat{k}\)) × (\(\hat{i}\) + λ \(\hat{j}\) + µ \(\hat{k}\)) = \(\overrightarrow{0}\)
हल:
प्रश्नानुसार,
(2 \(\hat{i}\) + 6 \(\hat{j}\) + 27 \(\hat{k}\)) × (\(\hat{i}\) + λ \(\hat{j}\) + µ \(\hat{k}\)) = \(\overrightarrow{0}\)
या
2 \(\hat{i}\) × (\(\hat{i}\) + λ \(\hat{j}\) + µ \(\hat{k}\)) + 6 \(\hat{j}\) × (\(\hat{i}\) + λ\(\hat{j}\) + µ \(\hat{k}\)) + 27 \(\hat{k}\) (\(\hat{i}\) + λ \(\hat{j}\) + µ \(\hat{k}\)) = \(\overrightarrow{0}\)
या
2( \(\hat{i}\) × \(\hat{i}\)) + 2 λ( \(\hat{i}\) × \(\hat{j}\)) + 2 µ(\(\hat{i}\) × \(\hat{k}\))
+ 6(\(\hat{j}\) × \(\hat{i}\)) + 6λ( \(\hat{j}\) × \(\hat{j}\)) + 6 µ(\(\hat{j}\) × \(\hat{k}\))
+ 27(\(\hat{k}\) × \(\hat{i}\)) + 27λ(\(\hat{k}\) × \(\hat{j}\)) + 27µ(\(\hat{k}\) × \(\hat{k}\)) = \(\overrightarrow{0}\)
या
\(\overrightarrow{0}\) + 2 λ \(\hat{k}\) – 2 μ \(\hat{j}\) – 6 \(\hat{k}\) + \(\overrightarrow{0}\) + 6 μ \(\hat{i}\)
+ 27 \(\hat{j}\) – 27λ \(\hat{i}\) + \(\overrightarrow{0}\) = \(\overrightarrow{0}\)
∴ \(\hat{i}\) × \(\hat{i}\) = \(\hat{j}\) × \(\hat{j}\) = \(\hat{k}\) × \(\hat{k}\) = \(\overrightarrow{0}\)
\( \hat{i}\) × \(\hat{j}\) = \(\hat{k}\), \(\hat{j}\) × \(\hat{i}\) = \(-\hat{k}\)
\(\hat{j}\) × \(\hat{k}\) = \(\hat{i}\), \(\hat{k}\) × \(\hat{j}\) = \(-\hat{i}\)
\(\hat{k}\) × \(\hat{i}\) = \(\hat{j}\), \(\hat{i}\) × \(\hat{k}\) = \(-\hat{j}\)
या
(2λ – 6) \(\hat{k}\) + (6μ – 27λ) \(\hat{i}\) + (27 – 2μ) \(\hat{j}\) = \(\overrightarrow{0}\)
अत: 2λ – 6 = 0
∴ λ = 3
तथा 27 – 2μ = 0
∴ = \(\frac{27}{2}\)
]अतः
λ = 3
तथा μ = \(\frac{27}{2}\)
प्रश्न 6.
दिया हुआ है कि \(\vec{a}\) \(\vec{b}\) = \(\overrightarrow{0}\) और \(\vec{a}\) × \(\vec{b}\) = \(\overrightarrow{0}\). सदिश \(\vec{a}\) और \(\vec{b}\) के बारे में आप क्या निष्कर्ष निकाल सकते हैं?
हल:
\(\vec{a}\) \(\vec{b}\) = \(\overrightarrow{0}\)
तब \(\vec{a}\) = \(\overrightarrow{0}\) या \(\vec{b}\) = \(\overrightarrow{0}\) या \(\vec{a}\) ⊥ \(\vec{b}\)
(सदिश \(\vec{a}\) सदिश \(\vec{b}\) पर लम्ब है)
अब \(\vec{a}\) × \(\vec{b}\) = \(\overrightarrow{0}\)
दोनों एक साथ सम्भव नहीं हैं।
अत: \(\vec{a}\) = \(\overrightarrow{0}\) या \(\vec{b}\) = \(\overrightarrow{0}\)
या
\(|\vec{a}|\) = 0 या \(|\vec{b}|\) = 0
प्रश्न 7.
मान लीजिए सदिश \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) क्रमशः
a1 \(\hat{i}\) + a2 \(\hat{j}\) + a3 \(\hat{k}\), b1 \(\hat{i}\) + b2 \(\hat{j}\) + b3 \(\hat{k}\), c1 \(\hat{i}\) + c2 \(\hat{j}\) + c3 \(\hat{k}\)
के रूप में दिए हुए हैं, तब दर्शाइए कि
\(\vec{a}\) × (\(\vec{b}\) + \(\vec{c}\)) = \(\vec{a}\) × \(\vec{b}\) + \(\vec{a}\) × \(\vec{c}\)
हल:
प्रश्न 8.
यदि \(\vec{a}\) = \(\overrightarrow{0}\) अथवा \(\vec{b}\) = \(\overrightarrow{0}\) तब \(\vec{a}\) × \(\vec{b}\) = \(\overrightarrow{0}\) होता है। क्या विलोम सत्य है ? उदाहरण सहित अपने उत्तर की पुष्टि कीजिए।
हल:
प्रश्नानुसार,
\(\vec{a}\) = \(\overrightarrow{0}\) तब \(|\vec{a}|\) = 0
∴ \(\vec{a}\) × \(\vec{b}\) = \(|\vec{a}|\) \(| \vec{b}|\) sin θ \(\hat{n}\)
जहाँ \(\vec{a}\) तथा \(\vec{b}\) के बीच्चोण θ है।
= 0 \(|\vec{b}|\) sin θ \(\hat{n}\) = \(\overrightarrow{0}\)
इसी प्रकार, जब \(\vec{b}\) = \(\overrightarrow{0}\)
तब \(|\vec{b}|\) = 0, \(\vec{a}\) × \(\vec{b}\) = \(\overrightarrow{0}\)
विलोम के लिए:
माना
\(\vec{a}\) = a1 \(\hat{i}\) + a2 \(\hat{j}\) + a3 \(\hat{k}\)
\(\vec{b}\) = λ(a1\(\hat{i}\) + a2\(\hat{j}\) + a3\(\hat{k}\))
तथा
अर्थात् \(\vec{a}\) तथा \(\vec{b}\) समान्तर हैं। अत:
\(|\vec{a}|\) ≠ 0, \(|\vec{b}|\) ≠ 0
\(\vec{a}\) × \(\vec{b}\) = \(|\vec{a}|\)\( |\vec{b}|\) sinθ \(\hat{n}\) = \(\overrightarrow{0}\)
अत:
\(\vec{a}\) × \(\vec{b}\) = \(\overrightarrow{0}\)
जब \(\vec{a}\) ≠ 0, \(\vec{b}\) ≠ \(\overrightarrow{0}\)
अत: विलोम सत्य नहीं है।
प्रश्न 9.
एक त्रिभुज का क्षेत्रफल ज्ञात कीजिए जिसके शीर्ष A(1, 1, 2), B(2, 3, 5) और C(1, 5, 5) हैं।
हल:
माना O मूलबिन्दु है, तब O के सापेक्ष ∆A B C में,
शीर्ष A का स्थिति सदिश = \(\overrightarrow{O A}\) = \(\hat{i}\) + \(\hat{j}\) + 2 \(\hat{k}\)
शीर्ष B का स्थिति सदिश = \(\overrightarrow{O B}\) = 2 \(\hat{i}\) + 3 \(\hat{j}\) + 5 \(\hat{k}\)
शीर्ष C का स्थिति सदिश = \(\overrightarrow{O C}\) = \(\hat{i}\) + 5 \(\hat{j}\) + 5 \(\hat{k}\)
अब
\(\overrightarrow{A B}\) = \(\overrightarrow{O B}\) – \(\overrightarrow{O A}\)
= (2 \(\hat{i}\) + 3 \(\hat{j}\) + 5 \(\hat{k}\)) – ( \(\hat{i}\) + \(\hat{j}\) + 2 \(\hat{k}\))
= (2 – 1) \(\hat{i}\) + (3 – 1) \(\hat{j}\) + (5 – 2) \(\hat{k}\)
= \(\hat{i}\) + 2 \(\hat{j}\) + 3 \(\hat{k}\)
\(\overrightarrow{A C}\) = \(\overrightarrow{O C}\) – \(\overrightarrow{O A}\)
= (\(\hat{i}\) + 5 \(\hat{j}\) + 5 \(\hat{k}\)) – (\(\hat{i}\) + \(\hat{j}\) + 2 \(\hat{k}\))
= (1 – 1) \(\hat{i}\) + (5 – 1) \(\hat{j}\) + (5 – 2) \(\hat{k}\)
= 0 + 4 \(\hat{j}\) + 3 \(\hat{k}\)
अत:
\(\overrightarrow{A B}\) × \(\overrightarrow{A C}\) = \(\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 0 & 4 & 3\end{array}\right|\)
= \(\left|\begin{array}{ll}
2 & 3 \\
4 & 3
\end{array}\right|\) \(\hat{i}\) – \(\left|\begin{array}{ll}
1 & 3 \\
0 & 3
\end{array}\right|\)\( \hat{j}\) + \(\left|\begin{array}{ll}
1 & 2 \\
0 & 4
\end{array}\right|\) \(\hat{k}\)
= (6 – 12) \(\hat{i}\) – (3 – 0) \(\hat{j}\) + (4 – 0) \(\hat{k}\)
= -6 \(\hat{i}\) – 3 \(\hat{j}\) + 4 \(\hat{k}\)
∴ \(|\overrightarrow{A B} ×\overrightarrow{A C}|\) = |-6 \(\hat{i}\) – 3 \(\hat{j}\) + 4 \(\hat{k}\)|
= \(\sqrt{(-6)^2+(-3)^2+4^2}\)
= \(\sqrt{36+9+16}\) = \(\sqrt{61}\)
∴ D A B C का क्षेत्रफल = \(\frac{1}{2}|\) \(\overrightarrow{A B}\) × \(\overrightarrow{A C}|\)
= \(\frac{1}{2}\) \(\sqrt{61}\) वर्ग इकाई।
प्रश्न 10.
एक समान्तर चतुर्भुज का क्षेत्रफल ज्ञात कीजिए जिसकी संलग्न भुजाएँ सदिश \(\vec{a}\) = \(\hat{i}\) – \(\hat{j}\) + 3 \(\hat{k}\) और \(\vec{b}\) = 2 \(\hat{i}\) – 7 \(\hat{j}\) + \(\hat{k}\) द्वारा निर्धारित हैं।
हल:
प्रश्नानुसार, समान्तर चतुर्भुज की संलग्न भुजाएँ हैं:
तथा
\(\vec{a}\) = \(\hat{i}\) – \(\hat{j}\) + 3 \(\hat{k}\)
तथा \(\vec{b}\) = 2 \(\hat{i}\) – 7 \(\hat{j}\) + \(\hat{k}\)
अत:
\(\vec{a}\) × \(\vec{b}\) = \(\left|\begin{array}{rrr}
\hat{i} & \hat{j} & \hat{k} \\
1 & -1 & 3 \\
2 & -7 & 1
\end{array}\right|\)
= \(\left|\begin{array}{ll}
-1 & 3 \\
-7 & 1
\end{array}\right|\)[\( \hat{i}\) – \(\left|\begin{array}{ll}
1 & 3 \\
2 & 1
\end{array}\right|\)\( \hat{j}\) + \(\left|\begin{array}{ll}
1 & -1 \\
2 & -7
\end{array}\right|\)\( \hat{k}\)
= [-1 × 1 – (-7) × 3] \(\hat{i}\) – [1 × 1 & -2 × 3] \(\hat{j}\)
+ [1 × (-7) – 2 × (-1)] \(\hat{k}\)
= (-1 + 21) \(\hat{i}\) – (1 – 6) \(\hat{j}\) + (-7 + 2) \(\hat{k}\)
= 20 \(\hat{i}\) + 5 \(\hat{j}\) – 5 \(\hat{k}\)
समान्तर चतुर्भुज का क्षेत्रफल
= \(|\vec{a} × \vec{b}|\)
= |20 \(\hat{i}\) + 5 \(\hat{j}\) – 5 \(\hat{k}\)|
= \(\sqrt{20^2+5^2+(-5)^2}\)
= \(\sqrt{400+25+25}\)
= \(\sqrt{450}\) = 15 \(\sqrt{2}\) वर्ग इकाई।
प्रश्न 11.
मान लीजिए सदिश \(\vec{a}\) और \(\vec{b}\) इस प्रकार हैं कि \(|\vec{a}|\) = 3 और \(|\vec{b}|\) = \(\frac{\sqrt{2}}{3}\), तब \(\vec{a}\) × \(\vec{b}\) एक मात्रक सदिश है यदि \(\vec{a}\) और \(\vec{b}\) के बीच का कोण है:
(A) \(\frac{\pi}{6}\)
(B) \(\frac{\pi}{4}\)
(C) \(\frac{\pi}{3}\)
(D) \(\frac{\pi}{2}\)
हल:
प्रश्नानुसार,
\(|\vec{a}|\) = 3 तथा \(|\vec{b}|\) = \(\frac{\sqrt{2}}{3}\)
तथा \(|\vec{a} × \vec{b}|\) = 1
अब \(\vec{a}\) × \(\vec{b}\) = \(|\vec{a}|\)\(|\vec{b}|\) sinθ \(\hat{n}\)
∴ \(|\vec{a} × \vec{b}|\) = \(|\vec{a}|\)\(|\vec{b}|\) sinθ \(|\hat{n}|\)
या 1 = \(|\vec{a}|\)\(|\vec{b}|\) sinθ (∴ \(|\hat{n}|\) = 1)
या 1 = 3 × \(\frac{\sqrt{2}}{3}\) sinθ
या sinθ = \(\frac{1}{\sqrt{2}}\) = sin \(\frac{\pi}{4}\)
∴ θ = \(\frac{\pi}{4}\)
अतः विकल्प B सही है।
प्रश्न 12.
एक आयत के शीर्ष A, B, C और D हैं, जिनके स्थिति सदिश क्रमशः
\(-\hat{i}\) + \(\frac{1}{2}\) \(\hat{j}\) + 4 \(\hat{k}\), \(\hat{i}\) + \(\frac{1}{2}\) \(\hat{j}\) + 4 \(\hat{k}\), \(\hat{i}\) – \(\frac{1}{2}\) \(\hat{j}\) + 4 \(\hat{k}\)
और \(-\hat{i}\) – \(\frac{1}{2}\) \(\hat{j}\) + 4 \(\hat{k}\) हैं, का क्षेत्रफल है:
(A) \(\frac{1}{2}\)
(B) 1
(C) 2
(D) 4
हल:
माना मूलबिन्दु O है, तब आयत A B C D के शीर्षों का मूलबिन्दु O के सापेक्ष
शीर्ष बिन्दु A का स्थिति सदिश
= \(\overrightarrow{O A}\) = \(-\hat{i}\) + \(\frac{1}{2}\) \(\hat{j}\) + 4 \(\hat{k}\)
शीर्ष बिन्दु B का स्थिति सदिश
= \(\overrightarrow{O B}\) = \(\hat{i}\) + \(\frac{1}{2}\) \(\hat{j}\) + 4 \(\hat{k}\)
शीर्ष बिन्दु C का स्थिति सदिश
= \(\overrightarrow{O C}\) = \(\hat{i}\) – \(\frac{1}{2}\) \(\hat{j}\) + 4 \(\hat{k}\)
शीर्ष बिन्दु D का स्थिति सदिश
= \(\overrightarrow{O D}\) = \(-\hat{i}\) \(-\frac{1}{2}\) \(\hat{j}\) + 4 \(\hat{k}\)
अब \(\overrightarrow{A B}\) = \(\overrightarrow{O B}\) – \(\overrightarrow{O A}\)
= ( \(\hat{i}\) + \(\frac{1}{2}\) \(\hat{j}\) + 4 \(\hat{k}\)) – (\(-\hat{i}\) + \(\frac{1}{2}\) \(\hat{j}\) + 4 \(\hat{k}\))
= (1 + 1) \(\hat{i}\) + (\(\frac{1}{2}\) – \(\frac{1}{2}\)) \(\hat{j}\) + (4 – 4) \(\hat{k}\)
= 2 \(\hat{i}\)
\(\overrightarrow{B C}\) = \(\overrightarrow{O C}\) – \(\overrightarrow{O B}\)
= (\(\hat{i}\) – \(\frac{1}{2}\)\(\hat{j}\) + 4 \(\hat{k}\)) – (\(\hat{i}\) + \(\frac{1}{i}\) + 4 \(\hat{k}\))
= \(-\hat{j}\)
∴ \(|\overrightarrow{A B}|\) = |2 \(\hat{i}\)| = 2,
\(|\overrightarrow{B C}|\) = \(|-\hat{j}|\) = 1
आयत का क्षेत्रफल = \(|\overrightarrow{A B}|\)\(|\overrightarrow{B C}|\)
= 2 × 1 = 2 वर्ग इकाई
अतः विकल्प C सही है।