al-Arbash Mohammad
𝑄1. (𝑎)
Determine whether the following three points lie on a straight line:
𝐴(1, −1,4), 𝐵 (−1,1,4) 𝑎𝑛𝑑 𝐶(1, 1, −1)
solution
Calculate the distance between these points
2
AB = √(−1 − 1)2 + (1 − (−1)) + (4 − 4)2 = 2√2
BC = √(1 − (−1))2 + (1 − 1)2 + (−1 − 4)2 = √29
2
AC = √(1 − 1)2 + (1 − (−1)) + (−1 − 4)2 = √29
𝐴𝐵 + BC ≠ AC , 𝐴𝐵 + AC ≠ BC , BC + AC ≠ AB
Non-collinear
𝑄1. (𝑏) Show that the equation x 2 + y 2 + 2y + z 2 + 8z = 0
𝑟𝑒𝑝𝑟𝑒𝑠𝑒𝑛𝑡 𝑎 𝑠𝑝ℎ𝑒𝑟𝑒, 𝑡ℎ𝑒𝑛 𝑓𝑖𝑛𝑑 𝑖𝑡𝑠 𝑐𝑒𝑛𝑡𝑒𝑟 𝑎𝑛𝑑 𝑟𝑎𝑑𝑖𝑢𝑠
solution
𝒙𝟐 + (𝒚𝟐 + 𝟐𝒚) + (𝒛𝟐 + 𝟖𝒛) = 𝟎
𝟐 𝟐
𝟖 𝟐
𝟐 𝟐
𝟖 𝟐
𝟐
𝒙 + (𝒚 + 𝟐𝒚 + ( ) ) + (𝒛 + 𝟖𝒛 + ( ) ) = 𝟎 + ( ) + ( )
𝟐
𝟐
𝟐
𝟐
𝟐
𝟐
𝒙𝟐 + (𝒚𝟐 + 𝟐𝒚 + 𝟏) + (𝒛𝟐 + 𝟖𝒛 + 𝟏𝟔) = 𝟎 + 𝟏 + 𝟏𝟔
𝒙𝟐 + (𝒚 + 𝟏)𝟐 + (𝒛 + 𝟒)𝟐 = 𝟏𝟕
𝑻𝒉𝒆 𝒄𝒆𝒏𝒕𝒆𝒓 𝒊𝒔 (𝟎, −𝟏, −𝟒)
𝑹𝒂𝒅𝒊𝒖𝒔 = √𝟏𝟕
al-Arbash Mohammad
𝑄2. (𝑎)
If A(4, −5) and B(1,4) are given two points, then find and
sketch the vector ⃗⃗⃗⃗⃗
BA. Is it a position vector? Explain your answer
solution
⃗⃗⃗⃗⃗
𝐵𝐴 = ⟨4 − 1, −5 − 4⟩ = ⟨3, −9⟩
⃗⃗⃗⃗⃗ = ⟨4 − 0, −5 − 0⟩ = ⟨4, −5⟩
OA
⃗⃗⃗⃗⃗
OB = ⟨1 − 0, 4 − 0⟩ = ⟨1, 4⟩
Thus, the position vector BA is equivalent to a vector that starts at the origin and is
directed to a point 3 units to the right along the x-axis and 9 units downward along
the y-axis.
al-Arbash Mohammad
𝑄2. (𝑏)
Are vectors 𝑣 = 4𝑖 − 5𝑘 and 𝑤 = −3𝑗 + 4𝑘 orthogonal to each other?
If not, find the angle between 𝑣 and 𝑤
solution
𝑣 = 4𝑖 − 5𝑘
, 𝑤 = −3𝑗 + 4𝑘
v ∙ w = 4(0) + 0(−3) + (−5)(4) = −20 ≠ 0
not orthogonal
𝑣. 𝑤
−20
𝑐𝑜𝑠𝜃 =
=
|𝑣||𝑤| √42 + (−5)2 ∙ √(−3)2 + (4)2
=
−20
√41 ∙ √25
=
−4
𝜃 = 𝑐𝑜𝑠 −1 (
) ≈ 128.66°
√41
−4
√41
al-Arbash Mohammad
𝑄3. (𝑎)
Find the area of the parallelogram with the vertices
𝑃(1,1, −1), 𝑄 (1, −1,4), 𝑅(4, −3, −5) 𝑎𝑛𝑑 𝑆(4, −1, −10)
solution
The area of a parallelogram is the magnitude of the cross-product of any
two non-parallel sides
𝑃𝑄 = ⟨1 − 1, −1 − 1, 4 − (−1)⟩ = ⟨0, −2, 5⟩
𝑃𝑅 = ⟨4 − 1, −3 − 1, −5 − (−1)⟩ = ⟨3, −4, −4⟩
𝑖
𝑃𝑄 × 𝑃𝑅 = |0
3
𝑗
−2
−4
𝑘
5 | = (8 + 20)𝑖 − (0 − 15)𝑗 + (0 + 6)𝑘
−4
= 28𝑖 + 15𝑗 + 6𝑘
Area = |𝑃𝑄 × 𝑃𝑅 | = √282 + 152 + 62 = √1045 ≈ 32.32645 …
al-Arbash Mohammad
𝑄3. (𝑏) Can you evaluate the area in part (a) by different vectors?
If yes give an example to corresponding vectors
Yes, we can
Because if we calculate the magnitude of the cross-product of any two non-parallel
sides, we will get the same area.
𝑃𝑄 = ⟨1 − 1, −1 − 1, 4 − (−1)⟩ = ⟨0, −2, 5⟩
𝑃𝑆 = ⟨4 − 1, −1 − 1, −10 − (−1)⟩ = ⟨3, −2, −9⟩
𝑖
𝑃𝑄 × 𝑃𝑆 = |0
3
𝑗
−2
−2
𝑘
5 | = (18 + 10)𝑖 − (0 − 15)𝑗 + (0 + 6)𝑘
−9
= 28𝑖 + 15𝑗 + 6𝑘
Area = |𝑃𝑄 × 𝑃𝑆| = √282 + 152 + 62 = √1045 ≈ 32.32645 …