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Ex 10.3, 13 (Method 1) If π β ,π β, π β are unit vectors such that π β + π β + π β = 0 β, find the value of π β .π β+π β . π β + π β. π β . Given π β ,π β, π β are unit vectors Magnitude of π β ,π β, π β is 1 So, |π β | = |π β | = |π β | = 1 Also, π β + π β + π β = 0 β So, |π β" + " π β" + " π β | = |π β | = 0 Now, |π β+π β+π β |2 = (π β + π β + π β) . (π β + π β + π β) = π β. π β + π β . π β + π β . π β + π β . π β + π β . π β + π β . π β + π β . π β + π β . π β + π β . π β = π β. π β + π β . π β + π β . π β + π β . π β + π β . π β + π β . π β + π β . π β + π β . π β + π β . π β = π β . π β + π β . π β + π β . π β + 2π β. π β + 2π β. π β + 2π β. π β = π β . π β + π β . π β + π β . π β + 2(π β. π β + π β. π β + π β. π β) = |π β |π + |π β |π + |π β |π + 2 (π β. π β + π β. π β + π β . π β) = 12 + 12 + 12 + 2(π β. π β + π β. π β + π β. π β) = 1 + 1 + 1 + 2(π β. π β + π β. π β + π β. π β) = 3 + 2 (π β. π β + π β. π β + π β. π β) β΄ |π β+π β+π β |2 = 3 + 2 (π β. π β + π β. π β + π β. π β) Now, |π β" + " π β" + " π β | = 0 |π β" + " π β" + " π β |^2 = 0 3 + 2 (π β. π β + π β. π β + π β. π β) = 0 2(π β. π β + π β. π β + π β. π β) = β3 (π β. π β + π β. π β + π β. π β) = (βπ)/π Ex 10.3, 13 (Method 2) If π β ,π β, π β are unit vectors such that π β + π β + π β = 0, find the value of π β .π β+π β . π β + π β. π β . Given π β ,π β, π β are unit vectors So, |π β | = |π β | = |π β | = 1 Also, ( π β + π β + π β ) = 0 β Now, π β . (π β + π β + π β) = π β . π β + π β. π β + π β . π β π β . 0 β = π β. π β + π β. π β + π β. π β 0 = π β. π β + π β. π β + π β. π β 0 β = |π β |π + π β. π β + π β. π β 0 β = |π β |π + π β. π β + π β. π β 0 β = |π β |2 + π β. π β + π β. π β 0 = 12 + π β. π β + π β. π β π β. π β + π β. π β = β1 Also, π β . (π β + π β + π β) = π β . π β + π β. π β + π β . π β π β . 0 β = π β. π β + π β. π β + π β. π β 0 = π β. π β + π β. π β + π β. π β 0 = π β. π β + π β. π β + π β. π β 0 = π β. π β + π β. π β + π β. π β 0 = π β. π β + |π β |2 + π β . π β 0 = π β. π β + 12 + π β . π β π β. π β + π β. π β = β1 Also π β . (π β + π β + π β) = π β . π β + π β . π β + π β . π β π β. 0 β = π β. π β + π β. π β + π β. π β 0 = π β. π β + π β. π β + π β. π β 0 = π β. π β + π β. π β + π β. π β 0 = π β. π β + π β. π β + π β. π β 0 = π β. π β + π β. π β + |π β |2 0 = π β. π β+ π β . π β + 12 π β. π β + π β. π β = β1 Adding (1), (2) and (3) (π β. π β + π β. π β) + (π β. π β + π β. π β) + (π β. π β + π β. π β) = β1 + (β1) + (β1) 2π β. π β + 2π β. π β + 2π β. π β = β3 2(π β. π β + π. π β + π β. π β) = β3 π β. π β + π β. π β + π β. π β = (βπ)/π Page 2
Last updated at April 21, 2021 by Teachoo
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Ex 10.3, 14 If either vector π β = 0 β or π β = 0 β, then π β. π β = 0 But the converse need not be true. justify your answer with an example. Converse: If π β . π β = 0, then either π β = 0 β or π β = 0 β Let π β = π Μ + π Μ + π Μ = 1π Μ + 1π Μ + 1π Μ and π β = π Μ + π Μ - 2π Μ = 1π Μ + 1π Μ β 2π Μ π β . π β = 1.1 + 1.1 + 1(β2) = 1 + 1 β 2 = 0 Hence, π β . π β = 0 but π β β 0 β and π β β 0 β Thus, the converse need not be true.
Updated April 25, 2017 By Ariel Balter, Ph.D.
To construct a vector that is perpendicular to another given vector, you can use techniques based on the dot-product and cross-product of vectors. The dot-product of the vectors A = (a1, a2, a3) and B = (b1, b2, b3) is equal to the sum of the products of the corresponding components: AβB = a1_b2 + a2_b2 + a3_b3. If two vectors are perpendicular, then their dot-product is equal to zero. The cross-product of two vectors is defined to be AΓB = (a2_b3 - a3_b2, a3_b1 - a1_b3, a1_b2 - a2*b1). The cross product of two non-parallel vectors is a vector that is perpendicular to both of them.
Write down a hypothetical, unknown vector V = (v1, v2).
Calculate the dot-product of this vector and the given vector. If you are given U = (-3,10), then the dot product is VβU = -3 v1 + 10 v2.
Set the dot-product equal to 0 and solve for one unknown component in terms of the other: v2 = (3/10) v1.
Pick any value for v1. For instance, let v1 = 1.
Solve for v2: v2 = 0.3. The vector V = (1,0.3) is perpendicular to U = (-3,10). If you chose v1 = -1, you would get the vector Vβ = (-1, -0.3), which points in the opposite direction of the first solution. These are the only two directions in the two-dimensional plane perpendicular to the given vector. You can scale the new vector to whatever magnitude you want. For instance, to make it a unit vector with magnitude 1, you would construct W = V/(magnitude of v) = V/(sqrt(10) = (1/sqrt(10), 0.3/sqrt(10).
Write down a hypothetical unknown vector V = (v1, v2, v3).
Calculate the dot-product of this vector and the given vector. If you are given U = (10, 4, -1), then VβU = 10 v1 + 4 v2 - v3.
Set the dot-product equal to zero. This is the equation for a plane in three dimensions. Any vector in that plane is perpendicular to U. Any set of three numbers that satisfies 10 v1 + 4 v2 - v3 = 0 will do.
Choose arbitrary values for v1 and v2, and solve for v3. Let v1 = 1 and v2 = 1. Then v3 = 10 + 4 = 14.
Perform the dot-product test to show that V is perpendicular to U: By the dot-product test, the vector V = (1, 1, 14) is perpendicular to the vector U: VβU = 10 + 4 - 14 = 0.
Choose any arbitrary vector that is not parallel to the given vector. If a vector Y is parallel to a vector X, then Y = a*X for some non-zero constant a. For simplicity, use one of the unit basis vectors, such as X = (1, 0, 0).
Calculate the cross product of X and U, using U = (10, 4, -1): W = XΓU = (0, 1, 4).
Check that W is perpendicular to U. WβU = 0 + 4 - 4 = 0. Using Y = (0, 1, 0) or Z = (0, 0, 1) would give different perpendicular vectors. They would all lie in the plane defined by the equation 10 v1 + 4 v2 - v3 = 0.
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If a b c are three unit vectors such that a+b+c=1 and a is perpendicular to b
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