2. Check the following measures as scalars and vectors:
(i) 10 kg
(ii) 2 meters north-west
(iii) 40°
(iv) 40 Watt
(v) 10–19 coulomb
(vi) 20 m/sec2
(i) 10 kg
(ii) 2 meters north-west
(iii) 40°
(iv) 40 Watt
(v) 10–19 coulomb
(vi) 20 m/sec2
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1 Answer
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(i) 10kg involves only magnitude. So, it is scalar quantity.
(ii) 2 meters north-west involves both magnitude and direction. So, it is vector quantity.
(iii) 400 involves only magnitude. So, it is scalar quantity.
(iv) 400 watts involves only magnitude. So, it is scalar quantity.
(v) 10-19 coulomb involves only magnitude. So, it is scalar quantity.
(vi) 20m/s-2 involves magnitude and direction. So, it is vector quantity.
Similar Questions for you
6.00
b·a = c·a
|a+b-c|² = |a|²+|b|²+|c|²+2(a·b - b·c - a·c)
= 4+16+16+2(a·b - 0 - a·b) = 36
⇒ |a+b-c| = 6
(a+3b). (7a-5b) = 7|a|² - 5ab + 21ab - 15|b|² = 7|a|²+16ab-15|b|²=0.
(a-4b). (7a-2b) = 7|a|² - 2ab - 28ab + 8|b|² = 7|a|²-30ab+8|b|²=0.
Subtracting: 46ab - 23|b|² = 0 ⇒ 2ab = |b|².
Substituting: 7|a|² + 8|b|² - 15|b|² = 0 ⇒ 7|a|² = 7|b|² ⇒ |a|=|b|.
cosθ = ab/ (|a|b|) = ab/|b|² = (1/2)|b|²/|b|² = 1/2.
θ = 60°.
a×b=c ⇒ a.c=0, b.c=0.
|c|² = |a|²|b|² - (a.b)² = (3)|b|² - 1. |c|=√2. So |b|²=1, |b|=1.
Projection of b on a×c.
a×c = a× (a×b) = (a.b)a - (a.a)b = a - 3b.
|a-3b|² = |a|²+9|b|²-6 (a.b) = 3+9-6 = 6.
l = |b. (a-3b)|/|a-3b| = | (a.b)-3|b|²|/√6 = |1-3|/√6 = 2/√6.
3l² = 3 (4/6) = 2.
|a × b|² + |a . b|² = |a|²|b|²
8² + (a . b)² = 2² * 5²
64 + (a . b)² = 100
(a . b)² = 36
a . b = 6 (since angle seems acute from options, but could be -6).
a = i + j + 2k
b = -i + 2j + 3k
a + b = 3j + 5k
a . b = -1 + 2 + 6 = 7
a × b = |i, j, k; 1, 2; -1, 2, 3| = -i - 5j + 3k
(a - b) × b) = (a × b) - (b × b) = a × b
(a × (a - b) × b) = a × (a × b) = (a . b)a - (a . a)b = 7a - 6b
. The expression becomes (a + b) × (7a - 6b) × b)
= (a + b) × (7 (a ×&n
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