3. Classify the following as scalar and vector quantities:
(i) Time period
(ii) Distance
(iii) Force
(iv) Velocity
(v) Work done
3. Classify the following as scalar and vector quantities:
(i) Time period
(ii) Distance
(iii) Force
(iv) Velocity
(v) Work done
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1 Answer
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(i) Time period involves only magnitude. So, it is scalar quantity.
(ii) Distance involves only magnitude. So, it is scalar quantity.
(iii) Force involves both magnitude and direction. So, it is vector quantity.
(iv) Velocity involves both magnitude and direction. So, it is vector quantity.
(v) Work done involves only magnitude. So, it is scalar quantity.
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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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