b a The Vector or x Product In C4 the scalar product was covered a . b = |a||b|cosq Pronounced a dot b a b Suppose the angle between two vectors a and b is The scalar product is written as a . b and is defined as The dot must NEVER be missed out. The result of the scalar product is a scalar quantity not a vector ! The scalar product is sometimes called the “dot” product.
Using this definition: i . i = j . j = k . k = 1 as the angle between these unit vectors is zero and cos0 = 1 i . j = 0 j . k = 0 i . k = 0 j . i = 0 k . j = 0 k . i = 0 So an answer is obtained when the components of the vector are in the same direction.
Notice that the scalar product uses the magnitudes, a and b, of the vectors as well as the angle between them, so, we get a different answer for: different size vectors at the same angle, 8 10 6 10
6 6 10 10 We also have different answers for the same size vectors at different angles. 6 10 6 10 The scalar product was defined so that the answer is unique to any 2 vectors. It will enable us easily to find the angle between the vectors, even in 3 dimensions.
2 3 -1 4 -3 -1 2 4 The Scalar Product of 2 Column Vectors Suppose and Then to form , there are quantities to multiply However, six of these are perpendicular. e.g. is along the x-axis and is along the y-axis. 2 4 The scalar product of these components is zero. The other three multiplications e.g 2 and 3 . . . involve parallel components so the angle between them is zero.
SUMMARY For the scalar product of 2 column vectors, e.g. and we multiply the “tops”, “middles” and “bottoms” and add the results. So,
e.g.1 Find the scalar product of the vectors and Solution:
Perpendicular Vectors If a.b = 0 Then either a = 0 or b = 0 or a = 0 or b = 0 are trivial cases as they mean the vector doesn’t exist. So, we must have The vectors are perpendicular.
Finding Angles between Vectors The scalar product can be rearranged to find the angle between the vectors. Notice how careful we must be with the lines under the vectors. The r.h.s. is the product of 2 vectors divided by the product of the 2 magnitudes of the vectors
e.g. Find the angle between and Solution: Tip: If at this stage you get zero, STOP. The vectors are perpendicular. ( 3s.f. )
When solving problems, we have to be careful to use the correct vectors. e.g. The triangle ABC is given by Find the cosine of angle ABC. Solution: We always sketch and label a triangle B ( any shape triangle will do ) C BUT the a and b of the formula are not the a and b of the question. A We need the vectors and
𝐀𝐁 =𝐁`𝐬 −𝐀`𝐬 𝐀𝐁𝐁𝐀 𝐑𝐮𝐥𝐞
Finding Angles between Lines e.g. a With lines instead of vectors, we have 2 possible angles. We usually give the acute angle. ( If the obtuse angle is found, subtract from 180 . ) We use the 2 direction vectors only since these define the angle.
e.g. Find the acute angle, a, between the lines Solution: and where
e.g. Find the acute angle, a, between the lines and Solution: where and
e.g. Find the acute angle, a, between the lines and Solution: where and
e.g. Find the acute angle, a, between the lines and Solution: where and (nearest degree) Autograph
The Vector Product The vector product is defined as a b = |a| |b|sin 𝐧 𝐧 is a unit vector perpendicular to both a and b. To determine the direction of 𝑛 use the right hand rule where finger 1 is a, finger 2 is b and the thumb is a b a b is out of the paper as the direction of 𝐧 is out a b is into the paper as the direction of 𝐧 is in
a b = |a||b|sin 𝐧 Using this definition: i i = j j = k k = as the angle between these unit vectors is zero and sin0 = 0 ij = jk = ki = ik = ji = kj = k i j –j –k –i
a b = (a1i + a2j + a3k) (b1i + b2j + b3k) a b = |a||b|sin 𝐧 a b = (a1i + a2j + a3k) (b1i + b2j + b3k) = a1b2k – a1b3j – a2b1k + a2b3i + a3b1j – a3b2i = (a2b3 – a3b2)i + (a3b1 – a1b3)j + (a1b2 – a2b1)k This result is on pg 4 of the formula book written as a column vector Link to applet
So the vector is perpendicular to Ex So the vector is perpendicular to This fact will be essential in understanding the equation of a plane Link to demo Link to worksheet