How do you find the angle between the vectors #u=<3, 2># and #v=<4, 0>#?
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1 Answer
Explanation:
The angle #theta# between two vectors #vec A# and #vec B# is related to the modulus (or magnitude) and scaler (or dot) product of #vec A# and #vec B# by the relationship: Postbox 6 0 5 – powerful and flexible email clients.
# vec A * vec B = |A| |B| cos theta #
By convention when we refer to the angle between vectors we choose the acute angle.
So for this problem, let the angle betwen #vecu# and #vecv# be #theta# then:
#vec u=<<3, 2>># and #vec v=<<4, 0>>#
The modulus is given by;
# |vec u| = |<<3, 2>>| = sqrt(3^2+2^2)=sqrt(9+4)=sqrt(13) #
# |vec v| = |<<4, 0>>| = sqrt(4^2+0^2)=sqrt(16)=4 #
And the scaler product is:
# vec u * vec v = <<3, 2>> * <<4, 0>>#
# = (3)(4) + (2)(0)#
# = (12)#
And so using # vec A * vec B = |A| |B| cos theta # we have:
# 12 = sqrt(13) * 4 * cos theta #
# :. cos theta = (12)/(4sqrt(13))#
# :. cos theta = 0.83205 .. #
# :. theta = 33.69006 °#
# :. theta = 33.7 °#(3sf)
Related questions
Solution:
The exponent of a number shows how many times the number is multiplied by itself.
(i) (30 × 4−1) × 22
According to the rules of exponents,
a0 = 1 and a−m = 1/am
(30 × 4−1) × 22
= (1 + 1/4) × 22
= [(4 + 1)/4] × 22
= (5/4) × 22
= 5/22 × 22 [Since 4 = 2 × 2 = 22]
= 5
(ii) (2−1 × 4−1) ÷ 2−2
According to the rules of exponents,
(am)n = amn, a−m = 1/am
(2−1 × 4−1) ÷ 2−2
= [2−1 × {(2)2}−1] ÷ 2−2[Since 4 = 2 × 2 = 22]
= (2−1 × 2−2) ÷ 2−2 [∵(am)n = amn]
= 2−3 ÷ 2−2 [∵am × an = am + n]
= 2−3−(−2) [∵am ÷ an = am−n]
= 2−3 + 2 = 2−1
= 1/2 [∵a−m = 1/am]
(iii) (1/2)−2 + (1/3)−2 + (1/4)−2
According to the rules of exponents,
(a/b)−m = (b/a)m
(1/2)−2 + (1/3)−2 + (1/4)−2
= (2/1)2 + (3/1)2 + (4/1)2
= (2)2 + (3)2 + (4)2
= 4 + 9 + 16 = 29
(iv) (3−1 +4−1 + 5−1)0
According to the rules of exponents,
a0 = 1 and a−m = 1/am
(3−1 + 4−1 + 5−1)0
= (1/3 + 1/4 + 1/5)0 [Since a−m = 1/am]
= 1 [Since a0 = 1]
(v) {(−2/3)−2}2
According to the rules of exponents,
a−m = 1/am and (a/b)m = am/bm
{(−2/3)−2}2
= {(3/−2)2}2 [Sincea−m = 1/am]
={32/(−2)2}2 [Since(a/b)m = am/bm]
=(9/4)2 = 81/16
Video Solution:
Luminar 3 2 0 4 25
Find the value of (i) (3⁰× 4⁻¹) × 2²(ii) (2⁻¹× 4⁻¹) ÷ 2⁻²(iii) (1/2)⁻²+ (1/3)⁻²+ (1/4)⁻²(iv) (3⁻¹+4⁻¹+ 5⁻¹)⁰ (v) {(−2/3)⁻²}²
Class 8 Maths NCERT Solutions - Chapter 12 Exercise 12.1 Question 3
Summary:
Luminar 3 2 0 4 7
The value of the following (i) (30 × 4−1) × 22 (ii) (2−1 × 4−1) ÷ 2−2 (iii) (1/2)−2 + (1/3)−2 + (1/4)−2 (iv) (3−1 +4−1 + 5−1)0 (v) {(−2/3)−2}2 are (i) 5 (ii) 1/2 (iii) 29 (iv) 1 and (v) 81/16 respectively.
