Let be a vector such that Then the projection of on the vector
Let be a vector such that Then the projection of on the vector
Option 1 -
Option 2 -
Option 3 -
Option 4 -
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1 Answer
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Correct Option - 1
Detailed Solution:
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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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