Video Transcript
If vector 𝐀 is equal to three 𝐢
plus two 𝐣 minus three 𝐤, vector 𝐁 is equal to two 𝐢 minus two 𝐣 minus 𝐤, and
vector 𝐂 is equal to two 𝐢 minus 𝐣 minus three 𝐤, find two 𝐂 minus the dot
product of two 𝐀 and two 𝐁 plus four 𝐂 multiplied by vector 𝐁.
The expression we’re trying to
calculate here is quite complicated. We will, therefore, begin by
calculating the expression inside the parentheses, two 𝐁 plus four 𝐂. When multiplying any vector by a
scalar or constant, we simply multiply each of the components by that scalar. This means that two 𝐁 is equal to
two multiplied by two 𝐢 minus two 𝐣 minus 𝐤. This simplifies to four 𝐢 minus
four 𝐣 minus two 𝐤.
We can calculate four 𝐂 in a
similar way. This is equal to eight 𝐢 minus
four 𝐣 minus 12𝐤. Our next step is to add these two
vectors. When adding any two vectors, we
simply add their corresponding components. Four 𝐢 plus eight 𝐢 is equal to
12𝐢, negative four 𝐣 plus negative four 𝐣 is equal to negative eight 𝐣, and
negative two 𝐤 plus negative 12𝐤 is negative 14𝐤. Two 𝐁 plus four 𝐂 is equal to
12𝐢 minus eight 𝐣 minus 14𝐤.
Our next step is to find a dot
product of two 𝐀 and two 𝐁 plus four 𝐂. Multiplying vector 𝐀 by two gives
us six 𝐢 plus four 𝐣 minus six 𝐤. We calculate the dot or scalar
product of any two vectors by multiplying the corresponding components and then
finding the sum of these values. We begin by multiplying 12 and
six. Next, we need to multiply negative
eight and four. And finally, we multiply negative
14 and negative six. The dot product is equal to 12
multiplied by six plus negative eight multiplied by four plus negative 14 multiplied
by negative six. This simplifies to 72 plus negative
32 plus 84, which in turn is equal to 124.
The entire expression inside the
square brackets which is the dot product of two 𝐀 and two 𝐁 plus four 𝐂 is equal
to the scalar or constant 124. Multiplying this by vector 𝐁, we
have that 124𝐁 is equal to 248𝐢 minus 248𝐣 minus 124𝐤. We can now work out the final
answer by subtracting this from two 𝐂. Two 𝐂 is equal to four 𝐢 minus
two 𝐣 minus six 𝐤. Subtracting the 𝐢-components gives
us negative 244𝐢. Subtracting the 𝐣-components gives
us positive 246𝐣. Finally, subtracting the
𝐤-components gives us 118𝐤.
If vector 𝐀 is equal to three 𝐢
plus two 𝐣 minus three 𝐤, vector 𝐁 is equal to two 𝐢 minus two 𝐣 minus 𝐤, and
vector 𝐂 is equal to two 𝐢 minus 𝐣 minus three 𝐤, then two 𝐂 minus the dot
product of two 𝐀 and two 𝐁 plus four 𝐂 multiplied by vector 𝐁 is equal to
negative 244𝐢 plus 246𝐣 plus 118𝐤.
