NCERT Solutions for Class 12 Maths Chapter 10 सदिश बीजगणित Ex 10.3

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NCERT Solutions for Class 12 Maths Chapter 10 सदिश बीजगणित Ex 10.3

प्रश्न 1.
दो सदिशों \(\vec{a}\) तथा \(\vec{b}\) के परिमाण क्रमशः \(\sqrt{3}\) एवं 2 हैं और \(\vec{a}\)\(\vec{b}\) = \(\sqrt{6}\) है, तो \(\vec{a}\) तथा \(\vec{b}\) के बीच का कोण ज्ञात कीजिए।
हल:
माना \(\vec{a}\) तथा \(\vec{b}\) के बीच का कोणθ है, तब
cosθ = \(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}\) = \(\frac{\sqrt{6}}{\sqrt{3} \times 2}\) = \(\frac{1}{\sqrt{2}}\)
∴ \(|\vec{a}|\) = \(\sqrt{3}\)
तथा
\(|\vec{b}|\) = 2
∴θ = cos^-1 \(\frac{1}{\sqrt{2}}\)
या θ = \(\frac{\pi}{4}\)
अतः सदिश्शो \(\vec{a}\) तथा \(\vec{b}\) के बीच का कोण \(\frac{\pi}{4}\) है।

प्रश्न 2.
सदिशों \(\hat{i}\) – 2 \(\hat{j}\) + 3 \(\hat{k}\) और 3\(\hat{i}\) – 2\(\hat{j}\) + \(\hat{k}\) के बीच का कोण ज्ञात कीजिए।
हल:
माना
\(\vec{a}\) = \(\hat{i}\) – 2\(\hat{j}\) + 3 \(\hat{k}\)
\(\vec{b}\) = 3 \(\hat{i}\) – 2\(\hat{j}\) + \(\hat{k}\)
तथा \(\vec{a}\) और \(\vec{b}\) केंगीचीच का कोण θ है।
तब cosθ = \(\frac{\vec{a} \cdot \vec{b}}{|\vec{a} \cdot| \vec{b} \mid}\)
अब \(\vec{a}\). \(\vec{b}\) = (\(\hat{i}\) – 2 \(\hat{j}\) + 3 \(\hat{k}\)) (3 \(\hat{i}\) – 2 \(\hat{j}\) + \(\hat{k}\))
[∴ ( a₁\(\hat { i }\) + a₂ \(\hat { j }\) + a₃ \(\hat { k }\) ) (b₁ \(\hat{i}\) + b₂ \(\hat{j}\) + b₃ \(\hat{k}\)
= a₁ b₁ + a₂ b₂ + a₃ b₃]
= 1 × 3 + (-2)(-2) + 3 × 1 = 3 + 4 + 3 = 10
∴ \(\vec{a}\) \(\vec{b}\) = 10
पुन:
\(|\vec{a}|\) = |\(\hat{i}\) – 2 \(\hat{j}\) + 3 \(\hat{k}\)|
= \(\sqrt{(1)^2+(-2)^2+3^2}\)
= \(\sqrt{1+4+9}\) = \(\sqrt{14}\)
\(|\vec{b}|\) = |3 \(\hat{i}\) – 2 \(\hat{j}\) + \(\hat{k}\)|
= \(\sqrt{3^2+(-2)^2+1^2}\) = \(\sqrt{9+4+1}\) = \(\sqrt{14}\)
\(\vec{a}\) \(\vec{b}\), \(|\vec{a}|\) तथा \(|\vec{b}|\) के समी. (1) में रखने पर,
cosθ = \(\frac{10}{\sqrt{14} \times \sqrt{14}}\) = \(\frac{10}{14}\) = \(\frac{5}{7}\)
∴θ = cos^-1 \((\frac{5}{7})\)

प्रश्न 3.
सदिश \(\hat{i}\) + \(\hat{j}\) पर सदिश \(\hat{i}\) – \(\hat{j}\) का प्रक्षेप ज्ञात कीजिए।
हल:
हम जानते हैं कि सदिश \(\vec{a}\) का सदिश \(\vec{b}\) पर प्रक्षेप
= \(\vec{a}\) \(\frac{\vec{b}}{|\vec{b}|}\)
माना
\(\vec{a}\) = \(\hat{i}\) + \(\hat{j}\)
तथा
\(\vec{b}\) = \(\hat{i}\) – \(\hat{j}\)
तब
\(\vec{b}\) = \(\hat{i}\) – \(\hat{j}\) का सदिश \(\vec{a}\) = \(\hat{i}\) + \(\hat{j}\) पर प्रक्षेप
= \(\vec{b}\) \(\frac{\vec{a}}{|\vec{a}|}\) = \(\frac{(\hat{i}-\hat{j}) (\hat{i}+\hat{j})}{|\hat{i}+\hat{j}|}\)
= \(\frac{\hat{i} \cdot \hat{i}+\hat{i} \cdot \hat{j}-\hat{j} \cdot \hat{i}-\hat{j} \cdot \hat{j}}{\sqrt{1^2+1^2}}\)
= \(\frac{|\hat{i}|^2+0-0-|\hat{j}|^2}{\sqrt{2}}\)
= \(\frac{1-1}{\sqrt{2}}\)
= 0

प्रश्न 4.
सदिश \(\hat{i}\) + 3 \(\hat{j}\) + 7 \(\hat{k}\) का सदिश 7\(\hat{i}\) – \(\hat{j}\) + 8 \(\hat{k}\) पर प्रक्षेप ज्ञात कीजिए।
हल:
माना \(\vec{a}\) = \(\hat{i}\) + 3 \(\hat{j}\) + 7 \(\hat{k}\)
तथा \(\vec{b}\) = 7 \(\hat{i}\) – \(\hat{j}\) + 8 \(\hat{k}\)
तब सदिश \(\vec{a}\) का सदिश \(\vec{b}\) पर प्रक्षेप
= \(\vec{a}\) \(\frac{\vec{b}}{|\vec{b}|}\)
= \(\frac{(\hat{i}+3 \hat{j}+7 \hat{k}) \cdot(7 \hat{i}-\hat{j}+8 \hat{k})}{|7 \hat{i}-\hat{j}+8 \hat{k}|}\)
= \(\frac{1 \times 7+3 \times(-1)+7 \times 8}{\sqrt{7^2+(-1)^2+8^2}}\)
= \(\frac{7-3+56}{\sqrt{49+1+64}}\) = \(\frac{60}{\sqrt{114}}\)

प्रश्न 5.
दर्शाइए कि दिए हुए निम्नलिखित तीन सदिशों में से प्रत्येक मात्रक सदिश है:
\(\frac{1}{7}\)(2 \(\hat{i}\) + 3 \(\hat{j}\) + 6 \(\hat{k}\)), \(\frac{1}{7}\)(3 \(\hat{i}\) – 6 \(\hat{j}\) + 2 \(\hat{k}\)), \(\frac{1}{7}\)(6 \(\hat{i}\) + 2 \(\hat{j}\) – 3\( \hat{k}\))
यह भी दर्शाइए कि ये सदिश परस्पर एक-दूसरे के लम्बवत् हैं।
हल:

प्रश्न 6.
यदि (\(\vec{a}\) + \(\vec{b}\)) (\(\vec{a}\) – \(\vec{b}\)) = 8 और \(|\vec{a}|\) = 8\(|\vec{b}|\) हो तो \(|\vec{a}|\) एवं \(|\vec{b}|\) ज्ञात कीजिए।
हल:
(\(\vec{a}\) + \(\vec{b}\)) (\(\vec{a}\) – \(\vec{b}\))
\(\vec{a}\)(\(\vec{a}\) – \(\vec{b}\)) + \(\vec{b}\)(\(\vec{a}\) – \(\vec{b}\))
= \(\vec{a}\) \(\vec{a}\) – \(\vec{a}\) \(\vec{b}\) + \(\vec{b}\)\(\vec{a}\) – \(\vec{b}\) \(\vec{b}\)
= \(|\vec{a}|^2\) – \(|\vec{b}|^2\) = 8 प्रश्नानुसार)
तब
\(|\vec{a}|^2\) – \(|\vec{b}|^2\) = 8 \(|\vec{a}|\) = 8\(|\vec{b}|\)
या
64\(|\vec{b}|^2\) – \(|\vec{b}|^2\) = 8
या
63\(|\vec{b}|^2\) = 8
∴ \(|\vec{b}|^2\) = \(\frac{8}{63}\)
या
\(|\vec{b}|\) = \(\sqrt{\frac{8}{63}}\) = \(\frac{2 \sqrt{2}}{3 \sqrt{7}}\)
परन्तु
\(|\vec{a}|\) = 8\(|\vec{b}|\) = 8 × \(\sqrt{\frac{8}{63}}\)
= \(\frac{8 \sqrt{8}}{\sqrt{63}}\) = \(\frac{8 \times 2 \sqrt{2}}{3 \sqrt{7}}\) = \(\frac{16 \sqrt{2}}{3 \sqrt{7}}\)
अत:
\(\vec{a}\) = \(\frac{16 \sqrt{2}}{3 \sqrt{7}}, \vec{b}\) = \(\frac{2 \sqrt{2}}{3 \sqrt{7}}\)

प्रश्न 7.
(3 \(\vec{a}\) – 5 \(\vec{b}\)) (2\(\vec{a}\) + 7 \(\vec{b}\)) का मान ज्ञात कीजिए।
हल:
(3 \(\vec{a}\) – 5 \(\vec{b}\)) (2 \(\vec{a}\) + 7 \(\vec{b}\))
= 3\(\vec{a}\) (2 \(\vec{a}\) + 7 \(\vec{b}\)) 5 \(\vec{b}\) (2 \(\vec{a}\) + 7 \(\vec{b}\))
= 6(\(\vec{a}\) \(\vec{a}\)) + 21 \(\vec{a}\) \(\vec{b}\) – 10 \(\vec{b}\) \(\vec{a}\) – 35 \(\vec{b}\) \(\vec{b}\)
= 6\(|\vec{a}|^2\) + 11 \(\vec{a}\) \(\vec{b}\) – 35\(|\vec{b}|^2\)
[ ∴ \(\vec{a}\) \(\vec{a}\) = \(|\vec{a}|^2 \) \(\vec{b}\) \(\vec{b}\) = \(|\vec{b}|^2\)]

प्रश्न 8.
दो सदिशों \(\vec{a}\) और \(\vec{b}\) के परिमाण ज्ञात कीजिए, यदि इनके परिमाण समान हैं और इनके बीच का कोण 60° है तथा इनका अदिश गुणनफल \(\frac{1}{2}\) है।
हल:
प्रश्नानुसार,
\(\vec{a}\) तथा \(\vec{b}\) के बीच का कोण θ = 60°
माना \(|\vec{a}|\) = \(|\vec{b}|\) = k
अब
cosθ =\(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}\) = \(\frac{\frac{1}{2}}{k \cdot k}\) = \(\frac{1}{2 k^2}\)
या cos 60° = \(\frac{1}{2 k^2}\)
या
\(\frac{1}{2}\) = \(\frac{1}{2 h^2}\)
k² = 1
k = pm
परंतु
k = \(|\vec{a}|\) = \(|\vec{b}|\)
∴ k = 1 = \(|\vec{a}|\) = \(|\vec{b}|\)
क्योंकि \(|\vec{a}|\) ≠ -1
तथा \(|\vec{a}|\) ≠ -1
∴ \(|\vec{a}|\) = 1 तथा \(|\vec{b}|\) = 1

प्रश्न 9.
यदि एक मात्रक सदिश \(\vec{a}\) वे लिए (\(\vec{x}\) – \(\vec{a}\)) (\(\vec{x}\) + \(\vec{a}\)) = 12 हो तो \(|\vec{x}|\) ज्ञात कीजिए।
हल:
प्रश्नानुसार,
\( \vec{a}|\) = 1
तथा
(\(\vec{x}\) – \(\vec{a}\)) (\(\vec{x}\) + \(\vec{a}\)) = 12
या
\(\vec{x}\) (\(\vec{x}\) + \(\vec{a}\)) – \(\vec{a}\) (\(\vec{x}\) + \(\vec{a}\)) = 12
या \(\vec{x}\) \(\vec{x}\) + \(\vec{x}\) \(\vec{a}\) – \(\vec{a}\) \(\vec{x}\) – \(\vec{a}\) \(\vec{a}\) = 12
या
\(|\vec{x}|^2\) – \(|\vec{a}|^2\) = 12 (∴\(\vec{x}\) \(\vec{a}\) = \(\vec{a}\) \(\vec{x}\))
या
\(|\vec{x}|^2\) – 1 = 12
या
\(|\vec{x}|^2\) = 12 + 1 = 13
या
\(|\vec{x}|\) = \(\sqrt{13}\)

प्रश्न 10.
यदि \(\vec{a}\) = 2 \(\hat{i}\) + 2 \(\hat{j}\) + 3\(\hat{k}\)
और
\(\vec{b}\) = \(-\hat{i}\) + 2 \(\hat{j}\) + \(\hat{k}\)
\(\vec{c}\) = 3 \(\hat{i}\) + \(\hat{j}\)
इस प्रकार हैं कि \(\vec{a}\) + λ \(\vec{b}\), \(\vec{c}\) पर लम्ब है, तो λ का मान ज्ञात कीजिए।
हल:
प्रश्नानुसार, \(\vec{a}\) + λ \(\vec{b}\), \(\vec{c}\) पर लम्ब है।
∴ (\(\vec{a}\) + λ \(\vec{b}\)) . \(\vec{c}\) = 0
{(2 \(\hat{i}\) + 2 \(\hat{j}\) + 3 \(\hat{k}\)) + λ(-\(\hat{i}\) + 2 \(\hat{j}\) + \(\hat{k}\))} (3 \(\hat{i}\) + \(\hat{j}\)) = 0
या {(2 – λ) \(\hat{i}\) + (2 + 2λ) \(\hat{j}\) + (3 + λ) \(\hat{k}\)} = 0
या
3(2 – λ) \(\hat{i}\) \(\hat{i}\) + (2 – λ)(\(\hat{i}\) \(\hat{j}\)) + 3(2 + 2 λ)(\(\hat{j}\) \(\hat{i}\))
+ (2 + 2λ) \(\hat{j}\) \(\hat{j}\) + 3(3 + λ) \(\hat{k}\) \(\hat{i}\) + (3 + λ) \(\hat{k}\) \(\hat{j}\) = 0
3(2 – λ) \(\hat{i}\)\(\hat{i}\) + (2 + 2λ) \(\hat{j}\) \(\hat{j}\) = 0
{[∴ \(\hat{i}\) \(\hat{j}\) = 0 = \(\hat{k}\) \(\hat{i}\) = \(\hat{j}\) \(\hat{k}\), \(\hat{i}\) \(\hat{i}\) = \(\hat{j}\) \(\hat{j}\) = 1]}
3(2 – λ) + (2 + 2 λ) = 0
6 – 3λ + 2 + 2λ = 0
8 – λ = 0
λ = 8

प्रश्न 11.
दर्शाइए कि दो शून्येतर सदिशों \(\vec{a}\) और \(\vec{b}\) के लिए \(|\vec{a}|\) \vec{b}[/latex] + \(|\vec{b}|\) \(\vec{a}\), \(|\vec{a}|\)\( \vec{b}\) – \(|\vec{b}|\)\( \vec{a}\) पर लम्ब है।
हल:
माना \(\vec{p}\) = \(|\vec{a}|\)\( \vec{b}\) + \(|\vec{b}|\)\( \vec{a}\)
तथा \(\vec{q}\) = \(|\vec{a}|\)\( \vec{b}\) – \(|\vec{b}|\)\( \vec{a}\)
अब \(\vec{p}\) \(\vec{q}\) ={\(|\vec{a}|\)\( \vec{b}\) + \(|\vec{b}|\)\( \vec{a}\)}.
{\(|\vec{a}|\)\( \vec{b}\) – \(|\vec{b}|\)\( \vec{a}\)}
= (\(|\vec{a}|\)\( \vec{b}\)) (\(|\vec{a}|\)\( \vec{b}\) – \(|\vec{b}|\)\( \vec{a}\)) + (\(|\vec{b}|\)\( \vec{a}\))
(\(|\vec{a}|\)\( \vec{b}\) – \(|\vec{b}|\)\( \vec{a}\))
= \(|\vec{a}|^2\) (\(\vec{b}\) \(\vec{b}\)) – \(|\vec{a}|\)\(|\vec{b}|\)( \(\vec{b}\) \(\vec{a}\))
+ \(|\vec{b}|\)\(|\vec{a}|\)(\(\vec{a}\) \(\vec{b}\)) – \(|\vec{b}|^2\) (\(\vec{a}\) \(\vec{a}\))
= \(|\vec{a}|^2\) \(|\vec{b}|^2\) + 0 – \(|\vec{b}|^2\) \(|\vec{a}|^2\)
∴ \( \vec{a}|\) \(| \vec{b}|\) = \(\vec{b}\) \(\vec{a}\)
तथा
\(\vec{a}\) \(\vec{b}\) = \(\vec{b}\) \(\vec{a}\)
अत: \(\vec{p}\) तथा \(\vec{q}\) परस्पर लम्ब हैं।
अर्थात् \(|\vec{a}|\)\( \vec{b}\) + \(|\vec{b}|\)\( \vec{a}\), \(|\vec{a}|\)\( \vec{b}\) – \(|\vec{b}|\)\( \vec{a}\) पर लम्ब है।

प्रश्न 12.
यदि \(\vec{a}\) \(\vec{a}\) = 0 और \(\vec{a}\) \(\vec{b}\) = 0 तो सदिश \(\vec{b}\) के बारे में क्या निष्कर्ष निकाला जा सकता है?
हल:
प्रश्नानुसार,
अब
\( \vec{a}\) \(\vec{a}\) = 0
तथा
\(\vec{a}\) \(\vec{b}\) = 0
\(\vec{a}\) \(\vec{a}\) = \(|\vec{a}|^2\) = 0
\(\vec{a}\) = 0
\(\vec{a}\) \(\vec{b}\) = \(|\vec{a}|\)\(|\vec{b}|\) cosθ = 0
(\(\vec{a}\) \(\vec{b}\) = 0)
अत: \(\vec{b}\) कोई भी सदिश हो सकता है।

प्रश्न 13.
यदि \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) मात्रक सदिश इस प्रकार हैं कि \(\vec{a}\)\( \vec{b}\) \(\overrightarrow{0}\) तो \(\vec{a}\) \(\vec{b}\) + \(\vec{b}\) \(\vec{c}\) + \(\vec{c}\) \(\vec{a}\) का मान ज्ञात कीजिए।
हल:

प्रश्न 14.
यदि \(\vec{a}\) = \(\overrightarrow{0}\) अथवा \(\vec{b}\) = \(\overrightarrow{0}\) तब \(\vec{a}\) \(\vec{b}\) = 0, परन्तु विलोम का सत्य होना आवश्यक नहीं है। एक उदाहरण द्वारा अपने उत्तर की पुष्टि कीजिए।
हल:
माना \(\vec{a}\) = 2 \(\hat{i}\) + 3 \(\hat{j}\) + 6 \(\hat{k}\)
तथा
\(|\vec{a}|\) = \(\sqrt{2^2+3^2+6^2}\)
तब
= \(\sqrt{4+9+36}\) = \(\sqrt{49}\) = 7
तथा
\(|\vec{b}|\) = \(\sqrt{3^2+(-6)^2+2^2}\)
= \(\sqrt{9+36+4}\) = \(\sqrt{49}\) = 7
∴ \(|\vec{a}|\) ≠ 0 तथा \(|\vec{b}|\) ≠ 0
अब
\(\vec{a}\) \(\vec{b}\) = (2 \(\hat{i}\) + 3 \(\hat{j}\) + 6 \(\hat{k}\)) (3 \(\hat{i}\) – 6 \(\hat{j}\) + 2 \(\hat{k}\))
= 2 × 3 + 3 ×(-6) + 6 × 2
⇒ \(\vec{a}\) \(\vec{b}\) = 6 – 18 + 12 = 0
[∴a₁ \(\hat{i}\) + a₂ \(\hat{j}\) + a₃ \(\hat{k}\)) (b₁ \(\hat{i}\) + b₂ \(\hat{j}\) + b₃ \(\hat{k}\))
= a₁ b₁ + a₂ b₂ + a₃ b₃]
अतः हम देखते हैं कि
\( \vec{a}\) \(\vec{b}\) = 0
परन्तु
\(|\vec{a}|\) ≠ 0
तथा
\(|\vec{b}|\) ≠ 0

प्रश्न 15.
यदि किसी त्रिभुज A B C के शीर्ष क्रमशः (1, 2, 3), (-1, 0, 0), (0, 1, 2) हैं तो ∠A B C का मान ज्ञात कीजिए। ∠A B C, सदिशों \(\overrightarrow{B A}\) एवं \(\overrightarrow{B C}\) के बीच का कोण है।
हल:

प्रश्न 16.
दर्शाइए कि बिन्दु A(1, 2, 7), B(2, 6, 3) और C(3, 10, -1) संरेख हैं।
हल:
माना मूलबिन्दु है, तब O के सापेक्ष
बिन्दु A का स्थिति सदिश = \(\overrightarrow{O A}\) = \(\hat{i}\) + 2 \(\hat{j}\) + 7 \(\hat{k}\)
बिन्दु B का स्थिति सदिश = \(\overrightarrow{O B}\) = 2 \(\hat{i}\) + 6 \(\hat{j}\) + 3 \(\hat{k}\)
बिन्दु C का स्थिति सदिश = \(\overrightarrow{O C}\) = 3 \(\hat{i}\) + 10 \(\hat{j}\) – \(\hat{k}\)
अब
\(\overrightarrow{A B}\) = \(\overrightarrow{O B}\) – \(\overrightarrow{O A}\)
= 2 \(\hat{i}\) + 6 \(\hat{j}\) + 3 \(\hat{k}\) – (\(\hat{i}\) + 2 \(\hat{j}\) + 7 \(\hat{k}\))
= (2 – 1) \(\hat{i}\) + (6 – 2) \(\hat{j}\) + (3 – 7) \(\hat{k}\)
∴ \(\overrightarrow{A B}\) = \(\hat{i}\) + 4 \(\hat{j}\) – 4 \(\hat{k}\)
तथा \(\overrightarrow{B C}\) = \(\overrightarrow{O C}\) – \(\overrightarrow{O B}\)
= (3 \(\hat{i}\) + 10 \(\hat{j}\) – \(\hat{k}\)) – (2 \(\hat{i}\) + 6 \(\hat{j}\) + 3 \(\hat{k}\))
= (3 – 2) \(\hat{i}\) + (10 – 6) \(\hat{j}\) + (-1 – 3) \(\hat{k}\)
या
\(\overrightarrow{B C}\) = \(\hat{i}\) + 4 \(\hat{j}\) – 4 \(\hat{k}\)
\(\overrightarrow{A B}\) = \(\hat{i}\) + 4 \(\hat{j}\) – 4 \(\hat{k}\) = \(\overrightarrow{B C}\)
अर्थात् \(\overrightarrow{A B}\) तथा \(\overrightarrow{B C}\) एक ही सदिश को प्रदर्शित करते हैं। अत: 1, B तथा C इस सदिश के ही बिन्दु हैं।
∴ बिन्दु A, B तथा C संरेख हैं।

प्रश्न 17.
दर्शाइए कि सदिश
2 \(\hat{i}\) – \(\hat{j}\) + \(\hat{k}\), \(\hat{i}\) – 3 \(\hat{j}\) – 5 \(\hat{k}\) और 3 \(\hat{i}\) – 4 \(\hat{j}\) – 4 \(\hat{k}\)
एक समकोण त्रिभुज के शीर्षों की रचना करते हैं।
हल:
माना त्रिभुज के शीर्ष A, B तथा C हैं और मूल- बिन्दु O है। तब O के सापेक्ष
शीर्ष बिन्दु A का स्थिति सदिश \(\overrightarrow{O A}\) = 2 \(\hat{i}\) – \(\hat{j}\) + \(\hat{k}\)
शीर्ष बिन्दु B का स्थिति सदिश \(\overrightarrow{O B}\) = \(\hat{i}\) – 3 \(\hat{j}\) – 5 \(\hat{k}\)
शीर्ष बिन्दु C का स्थिति सदिश \(\overrightarrow{O C}\) = 3 \(\hat{i}\) – 4 \(\hat{j}\) – 4 \(\hat{k}\)
अब \(\overrightarrow{A B}\) = \(\overrightarrow{O B}\) – \(\overrightarrow{O A}\)
= (\(\hat{i}\) – 3 \(\hat{j}\) – 5 \(\hat{k}\)) – (2 \(\hat{i}\) – \(\hat{j}\) + \(\hat{k}\))
(i. 2) \(\hat{i}\) + (-3 + 1) \(\hat{j}\) + (-5 – 1) \(\hat{k}\)
या \(\overrightarrow{A B}\) = \(-\hat{i}\) – 2 \(\hat{j}\) – 6 \(\hat{k}\)
\(\overrightarrow{B C}\) = \(\overrightarrow{O C}\) – \(\overrightarrow{O B}\)
= (3 \(\hat{i}\) – 4 \(\hat{j}\) – 4 \(\hat{k}\)) – (\(\hat{i}\) – 3 \(\hat{j}\) – 5 \(\hat{k}\))
= (3 – 1) \(\hat{i}\) + (-4 + 3) \(\hat{j}\) + (-4 + 5) \(\hat{k}\)
या \(\overrightarrow{B C}\) = 2 \(\hat{i}\) – \(\hat{j}\) + \(\hat{k}\)
\(\overrightarrow{C A}\) = \(\overrightarrow{O A}\) – \(\overrightarrow{O C}\)
= (2 \(\hat{i}\) – \(\hat{j}\) + \(\hat{k}\)) – (3 \(\hat{i}\) – 4 \(\hat{j}\) – 4 \(\hat{k}\))
= (2 – 3) \(\hat{i}\) + (-1 + 4) \(\hat{j}\) + (1 + 4) \(\hat{k}\)
या
\(\overrightarrow{C A} = + 3 [latex]\hat{j}\) + 5 \(\hat{k}\)
पुन:
\((\overrightarrow{B C})\) \((\overrightarrow{C A})\)
= (2 \(\hat{i}\) – \(\hat{i}\) + \(\hat{k}\) (-\(\hat{i}\) + 3 \(\hat{j}\) + 5 \(\hat{k}\))
= 2 × (-1) + (-1) × 3 + 1 × 5
= -2 – 3 + 5 = 0
{\(\vec{a}\) \(\vec{b}\) = (a₁ b₁ + a₂ b₂ + a₃ b₃). से}
अतः B C ⊥ C A
∴ ∆A B C एक समकोण त्रिभुज है।
अतः दिए गए शीर्ष एक समकोण त्रिभुज की रचना करते हैं।

प्रश्न 18.
यदि शून्येतर सदिश \(\vec{a}\) का परिमाण ‘ a ‘ है और λ एक शून्येतर अदिश है तो λ \(\vec{a}\) एक मात्रक सदिश है, यदि:
(A) λ=1
(B) λ=-1
(C) a=|λ|
(D) a= \(\frac{1}{|\lambda|}\)
हल:
λ \(\vec{a}\) एक मात्रक सदिश है, तो
|λ \(\vec{a}\)| = 1
या
|λ|\(|\vec{a}|\) = 1
\(|\vec{a}|\) = \(\frac{1}{|\lambda|}\)
या
a = \(\frac{1}{|\lambda|}\) (∵ \(|\vec{a}|\) = a)
अतः विकल्प (D) सही है।