These NCERT Solutions for Class 12 Maths Chapter 10 सदिश बीजगणित Ex 10.5 Questions and Answers are prepared by our highly skilled subject experts.
NCERT Solutions for Class 12 Maths Chapter 10 सदिश बीजगणित Ex 10.5
प्रश्न 1.
यदि \(\vec{a}\) = \(\hat{i}\) – 2 \(\hat{j}\) + 3 \(\hat{k}\), \(\vec{b}\) = 2 \(\hat{i}\) – 3 \(\hat{j}\) + 4 \(\hat{k}\) और \(\vec{c}\) = 3 \(\hat{i}\) + \(\hat{j}\) – 2 \(\hat{k}\) है, तो [\(\vec{a}\) \(\vec{b}\) \(\vec{c}\)] ज्ञात कीजिए।
हल:
\(\vec{a}\) = \(\hat{i}\) – 2 \(\hat{j}\) + 3 \(\hat{k}\), \(\vec{b}\) = 2 \(\hat{i}\) – 3 \(\hat{j}\) + 4 \(\hat{k}\), \(\vec{c}\) = 3 \(\hat{i}\) + \(\hat{j}\) – 2 \(\hat{k}\)
{[\(\vec{a}\)\( \vec{b}\)\( \vec{c}\)] = \(\left[\begin{array}{ccc}
1 & -2 & 3 \\
2 & -3 & 1 \\
3 & 1 & -2
\end{array}\right]\)
= 1(6 – 1) + 2(-4 – 3) + 3(2 + 9)
= 5 – 14 + 33 = 33 – 9 = 24
प्रश्न 2.
दर्शाइए कि \(\vec{a}\) = \(\hat{i}\) – 2 \(\hat{j}\) + 3 \(\hat{k}\), \(\vec{b}\) = -2 \(\hat{i}\) + 3 \(\hat{j}\) – 4 \(\hat{k}\) और \(\vec{c}\) = \(\hat{i}\) – 3 \(\hat{j}\) + 5 \(\hat{k}\) सदिश समतलीय हैं।
हल:
\(\vec{a}\) = \(\hat{i}\) – 2 \(\hat{j}\) + 3 \(\hat{k}\), \(\vec{b}\) = -2 \(\hat{i}\) + 3 \(\hat{j}\) – 4 \(\hat{k}\), \(\vec{c}\) = \(\hat{i}\) – 3 \(\hat{j}\) + 5 \(\hat{k}\)
[\(\vec{a}\) \(\vec{b}\) \(\vec{c}\)] = \(\left[\begin{array}{ccc}
1 & -2 & 3 \\
-2 & 3 & -4 \\
1 & -3 & 5
\end{array}\right]\)
= 1(15 – 12) + 2(-10 + 4) + 3(6 – 3)
= 3 – 12 + 9 = 12 – 12 = 0 इति सिद्धम्
इति सिद्धम्
प्रश्न 3.
यदि सदिश \(\hat{i}\) – \(\hat{j}\) + \(\hat{k}\), 3 \(\hat{i}\) + \(\hat{i}\) + 2 \(\hat{k}\) और \(\hat{i}\) + λ \(\hat{j}\) – 3 \(\hat{k}\) समतलीय हैं, तो λ का मान ज्ञात कीजिए।
हल:
\({\left[\begin{array}{ccc}
1 & -1 & 1 \\
3 & 1 & 2 \\
1 & λ & -3
\end{array}\right]=0}\)
1(-3 – 2λ) + 1(-9 – 2) + 1(3λ – 1) = 0
-3 – 2 λ – 11 + 3λ – 1 = 0
λ = 15
प्रश्न 4.
मान लीजिए कि \(\vec{a}\) = \(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\), \(\vec{b}\) = \(\hat{i}\) और \(\bar{c}\) = c1 \(\hat{i}\) + c2 \(\hat{j}\) + c3 \(\hat{k}\) है। तब,
(a) यदि c1 = 1 और c2 = 2 है, तो c3 ज्ञात कीजिए, जिससे \(\vec{a}\), \(\vec{b}\) और \(\vec{c}\) समतलीय हो जाएँ।
(b) यदि c2 = -1 और c3 = 1 है, तो दर्शाइए कि c1 का कोई भी मान \(\vec{a}\), \(\vec{b}\) और \(\vec{c}\) को समतलीय नहीं बना सकता है।
(a) c1 = 1, c2 = 2, c3 = ?
\({\left[\begin{array}{lll}
1 & 1 & 1 \\
1 & 0 & 0 \\
1 & 2 & c_3
\end{array}\right]=0} \)
⇒ 1(c3 – 2) = 0 ⇒ c3 – 2
(b) c2 = -1, c3 = 2, c1 = ?
\(\left[\begin{array}{ccc}
1 & 1 & 1 \\
1 & 0 & 0 \\
c_1 & -1 & 2
\end{array}\right]=0\)
⇒ 1(0 – 0) – 1(2 – 0) + 1(- 1 – 0) = 0
⇒ 0 – 2 – 1 = 0
प्रश्न 5.
दर्शाइए कि स्थिति सदिशों 4 \(\hat{i}\) + 8 \(\hat{j}\) + 12 \(\hat{k}\), 2 \(\hat{i}\) + 4 \(\hat{j}\) + 6 \(\hat{k}\), 3 \(\hat{i}\) + 5 \(\hat{j}\) + 4 \(\hat{k}\) और 5 \(\hat{i}\) + 8 \(\hat{j}\) + 5 \(\hat{k}\) वाले चारों बिंदु समतलीय हैं।
हल:
माना \(\mathrm{A}\) = 4 \(\hat{i}\) + 8 \(\hat{j}\) + 12 \(\hat{k}\), B = -2 \(\hat{i}\) + 4 \(\hat{j}\) + 6 \(\hat{k}\)
C = 3 \(\hat{i}\) + 5 \(\hat{j}\) + 4 \(\hat{k}\), D = 5 \(\hat{i}\) + 8 \(\hat{j}\) + 5 \(\hat{k}\)
\(\overrightarrow{\mathrm{AB}}\) = (2 – 4) \(\hat{i}\) + (4 – 8) \(\hat{j}\) + (6 – 12) \(\hat{k}\) = -2 \(\hat{i}\) – 4 \(\hat{j}\) – 6 \(\hat{k}\)
\(\overrightarrow{\mathrm{BC}}\) = (3 – 2) \(\hat{i}\) + (5 – 4) \(\hat{j}\) + (4 – 6) \(\hat{k}\) = \(\hat{i}\) + \(\hat{j}\) – 2 \(\hat{k}\)
\(\overline{\mathrm{CD}}\) = (5 – 3) \(\hat{i}\) + (8 – 5) \(\hat{j}\) + (5 – 4) \(\hat{k}\)
= 2 \(\hat{i}\) + 3 \(\hat{j}\) + \(\hat{k}\)
\({\left[\begin{array}{ccc}
-2 & -4 & -6 \\
1 & 1 & -2 \\
2 & 3 & 1
\end{array}\right]}\)
= -2(1 + 6) + 4(1 + 4) -6 (3 – 2)
= -14 + 20 – 6 = 0
प्रश्न 6.
यदि चार बिंदु A(3, 2, 1), B(4, x, 5), C(4, 2, -2) और D(6, 5, -1) समतलीय हैं, तो x का मान ज्ञात कीजिए।
हल:
माना
\(\overrightarrow{\mathrm{OA}}\) = 3 \(\hat{i}\) + 2 \(\hat{j}\) + \(\hat{k}\); \(\overrightarrow{\mathrm{OB}}\) = 4 \(\hat{i}\) + x \(\hat{j}\) + 5 \(\hat{k}\)
\(\overrightarrow{\mathrm{OC}}\) = 4 \(\hat{i}\) + 2 \(\hat{j}\) – 2 \(\hat{k}\); \(\overrightarrow{\mathrm{OD}}\) = 6 \(\hat{i}\) + 5 \(\hat{j}\) – \(\hat{k}\)
∴ \(\overrightarrow{\mathrm{AB}}\) = \(\overrightarrow{\mathrm{OB}}\) – \(\overrightarrow{\mathrm{OA}}\) = \(\hat{i}\) + (x – 2) \(\hat{j}\) + 4 \(\hat{k}\)
\(\overrightarrow{\mathrm{AC}}\) = \(\overrightarrow{\mathrm{OC}}\) – \(\overrightarrow{\mathrm{OA}}\) = \(\hat{i}\) + 0 \(\hat{j}\) – 3 \(\hat{k}\)
\(\overrightarrow{\mathrm{AD}}\) = \(\overrightarrow{\mathrm{OD}}\) – \(\overrightarrow{\mathrm{OA}}\) = 3 \(\hat{i}\) + 3 \(\hat{j}\) – 2 \(\hat{k}\)
समतलीय होने की शर्तानुसार,
⇒\(\overrightarrow{\mathrm{AB}}\) \(\overrightarrow{\mathrm{AC}}\) \(\overrightarrow{\mathrm{AD}}\) = 0
⇒ \(\left[\begin{array}{ccc}
1 & x-4 & 4 \\
1 & 0 & -3 \\
3 & 3 & -2
\end{array}\right]=0\)
⇒ 1(0 + 9) -( x – 2)(-2 + 9) + 4(3 – 0) = 0
⇒ 9 – (x – 2)(7) + 12 = 0
⇒ -7 x + 14 + 21 = 0
⇒ -7 x = -35
⇒ x = 5
प्रश्न 7.
यदि \(\vec{a}\) + \(\vec{b}\), \(\vec{b}\) + \(\vec{c}\) और \(\vec{c}\) + \(\vec{a}\) समतलीय हैं, तो दर्शाइए कि सदिश \(\vec{a}\), \(\vec{b}\) और \(\vec{c}\) समतलीय होंगे।
हल:
[\(\vec{a}\) + \(\vec{b}\) \(\vec{b}\) + \(\vec{c}\) \(\vec{c}\) + \(\vec{a}\)] = 0
(\(\vec{b}\) + \(\vec{c}\)) × (\(\vec{c}\) + \(\vec{a}\)) = \(\vec{b}\) × (\(\vec{c}\) + \(\vec{a}\)) + \(\vec{c}\) × (\(\vec{c}\) + \(\vec{a}\))
= (\(\vec{b}\) × \(\vec{c}\)) + (\(\vec{b}\) × \(\vec{a}\)) + (\(\vec{c}\) × \(\vec{c}\)) + (\(\vec{c}\) × \(\vec{a}\))
= (\(\vec{b}\) × \(\vec{c}\)) + (\(\vec{b}\) × \(\vec{a}\)) + (\(\vec{c}\) × \(\vec{a}\))
= {[\(\vec{a}\) + \(\vec{b}\) \(\vec{b}\) + \(\vec{c}\) \(\vec{c}\) + \(\vec{a}\)] }
= (\(\vec{a}\) + \(\vec{b}\)) {(\(\vec{b}\) + \(\vec{c}\)) × (\(\vec{c}\) + \(\vec{a}\))}
= (\(\vec{a}\) + \(\vec{b}\)) {(\(\vec{b}\) × \(\vec{c}\)) + (\(\vec{b}\) × \(\vec{a}\)) + (\(\vec{c}\) × \(\vec{a}\))}
= (\(\vec{a}\) + \(\vec{b}\)) (\(\vec{b}\) × \(\vec{c}\)) + (\(\vec{a}\) + \(\vec{b}\)) (\(\vec{b}\) × \(\vec{a}\)) + (\(\vec{a}\) + \(\vec{b}\)) (\(\vec{c}\) × \(\vec{a}\))
= \(\vec{a}\) (\(\vec{b}\) × \(\vec{c}\)) + \(\vec{b}\) (\(\vec{b}\) × \(\vec{c}\)) + \(\vec{a}\) (\(\vec{b}\) × \(\vec{a}\)) = 0]
= {[\(\vec{a}\) \(\vec{b}\) \(\vec{c}\)] + 0 + 0 + 0 + 0 + [\(\vec{b}\) \(\vec{c}\) \(\vec{a}\)]}
= 2[\(\vec{a}\) \(\vec{b}\) \(\vec{c}\)]
∴ {[\(\vec{a}\) + \(\vec{b}\) \(\vec{b}\) + \(\vec{c}\) \(\vec{c}\) + \(\vec{a}\)] = 2[\(\vec{a}\) \(\vec{b}\) \(\vec{c}\)] (\(\vec{a}\)) + \(\vec{a}\) (\(\vec{c}\) × \(\vec{a}\)) + \(\vec{b}\) (\(\vec{c}\) × \(\vec{a}\)) }
⇒ {[\(\vec{a}\), \(\vec{b}\), \(\vec{c}\)] = 0 इति सिद्धम्।