What is the position vector of D D D A B C D ABCD A B C D
Question
The position vectors of points
A A A
,
B B B
and
C C C
are
a = 4 i − 9 j − k , b = i + 3 j + 5 k , and c = p i − j + 3 k . \begin{equation*} \mathbf{a} = 4\mathbf{i} - 9\mathbf{j} - \mathbf{k}, \quad \mathbf{b} = \mathbf{i} + 3\mathbf{j} + 5\mathbf{k}, \quad\text{and}\quad \mathbf{c} = p\mathbf{i} - \mathbf{j} + 3\mathbf{k}. \end{equation*} a = 4 i − 9 j − k , b = i + 3 j + 5 k , and c = p i − j + 3 k .
Find the unit vector parallel to the vector A B \mathbf{AB} AB
Find the value of p p p A A A B B B C C C
If p = − 2 p = -2 p = − 2 D D D A B C D ABCD A B C D
Solution
The vector A B → \overrightarrow{AB} A B A A A B B B
From the diagram, we have that
a + A B → = b \mathbf{a} + \overrightarrow{AB} = \mathbf{b} a + A B = b
, i.e. that
A B → = b − a = ( i + 3 j + 5 k ) − ( 4 i − 9 j − k ) = − 3 i + 12 j + 6 k . \begin{equation*} \overrightarrow{AB} = \mathbf{b} - \mathbf{a} = (\mathbf{i} + 3\mathbf{j} + 5\mathbf{k}) - (4\mathbf{i} - 9\mathbf{j} - \mathbf{k}) = -3 \mathbf{i} + 12 \mathbf{j} + 6 \mathbf{k}. \end{equation*} A B = b − a = ( i + 3 j + 5 k ) − ( 4 i − 9 j − k ) = − 3 i + 12 j + 6 k . The length of this vector is
( − 3 ) 2 + 12 2 + 6 2 = 9 + 144 + 36 = 189 = 3 21 , \begin{equation*} \sqrt{(-3)^2 + 12^2 + 6^2} = \sqrt{9 + 144 + 36} = \sqrt{189} = 3\sqrt{21}, \end{equation*} ( − 3 ) 2 + 1 2 2 + 6 2 = 9 + 144 + 36 = 189 = 3 21 , and therefore the unit vector parallel to
A B → \overrightarrow{AB} A B
is
− 3 3 21 i + 12 3 21 j + 6 3 21 k = − 1 21 i + 4 21 j + 2 21 k . \begin{align*} &\frac{-3}{3\sqrt{21}} \mathbf{i} + \frac{12}{3\sqrt{21}} \mathbf{j} + \frac{6}{3\sqrt{21}} \mathbf{k} \\ =&\frac{-1}{\sqrt{21}} \mathbf{i} + \frac{4}{\sqrt{21}} \mathbf{j} + \frac{2}{\sqrt{21}} \mathbf{k}. \end{align*} = 3 21 − 3 i + 3 21 12 j + 3 21 6 k 21 − 1 i + 21 4 j + 21 2 k . The points A , B A, B A , B C C C ⟺ A B → \iff \overrightarrow{AB} ⟺ A B A C → ⟺ A B → = k A C → \overrightarrow{AC} \iff \overrightarrow{AB} = k\overrightarrow{AC} A C ⟺ A B = k A C k k k
Now
A B → = − 3 i + 12 j + 6 k , A C → = ( p − 4 ) i + 8 j + 4 k , \overrightarrow{AB} = -3 \mathbf{i} + 12 \mathbf{j} + 6 \mathbf{k},\quad \overrightarrow{AC} = (p-4) \mathbf{i} + 8 \mathbf{j} + 4 \mathbf{k}, A B = − 3 i + 12 j + 6 k , A C = ( p − 4 ) i + 8 j + 4 k , and thus
2 A B → = − 6 i + 24 j + 12 k , 3 A C → = 3 ( p − 4 ) i + 24 j + 12 k . 2\overrightarrow{AB} = -6 \mathbf{i} + 24 \mathbf{j} + 12 \mathbf{k},\quad 3\overrightarrow{AC} = 3(p-4) \mathbf{i} + 24 \mathbf{j} + 12 \mathbf{k}. 2 A B = − 6 i + 24 j + 12 k , 3 A C = 3 ( p − 4 ) i + 24 j + 12 k . Comparing coefficients, we can say that A , B A, B A , B C C C − 6 = 3 ( p − 4 ) -6 = 3(p-4) − 6 = 3 ( p − 4 ) p = 2 p = 2 p = 2
Alternatively, we could you use the fact that C C C A B AB A B A B AB A B C C C p p p
We have the following sketch.
The position vector of
D D D
is therefore
d = a + A D → = a + B C → = a + c − b = ( 4 i − 9 j − k ) + ( − 2 i − j + 3 k ) − ( i + 3 j + 5 k ) = ( 4 − 2 − 1 ) i + ( − 9 − 1 − 3 ) j + ( − 1 + 3 − 5 ) k = i − 13 j − 3 k . \begin{align*} \mathbf{d} = \mathbf{a} + \overrightarrow{AD} = \mathbf{a} + \overrightarrow{BC}&= \mathbf{a} + \mathbf{c} - \mathbf{b} \\ &= \left( 4\mathbf{i} - 9\mathbf{j} - \mathbf{k} \right) + \left( -2\mathbf{i} - \mathbf{j} + 3\mathbf{k} \right) - \left( \mathbf{i} + 3\mathbf{j} + 5\mathbf{k} \right) \\ &= (4 - 2 - 1) \mathbf{i} + (-9 - 1 - 3) \mathbf{j} + (-1 + 3 - 5) \mathbf{k} \\ &= \mathbf{i} - 13 \mathbf{j} - 3 \mathbf{k}. \end{align*} d = a + A D = a + B C = a + c − b = ( 4 i − 9 j − k ) + ( − 2 i − j + 3 k ) − ( i + 3 j + 5 k ) = ( 4 − 2 − 1 ) i + ( − 9 − 1 − 3 ) j + ( − 1 + 3 − 5 ) k = i − 13 j − 3 k . Check: does
D C → = A B → \overrightarrow{DC} = \overrightarrow{AB} D C = A B
? We have that
A B → = b − a = ( i + 3 j + 5 k ) − ( 4 i − 9 j − k ) = − 3 i + 12 j + 6 k , \begin{align*} \overrightarrow{AB} &= \mathbf{b} - \mathbf{a} \\ &= \left( \mathbf{i} + 3\mathbf{j} + 5\mathbf{k} \right) - \left( 4\mathbf{i} - 9\mathbf{j} - \mathbf{k} \right) \\ &= -3 \mathbf{i} + 12 \mathbf{j} + 6 \mathbf{k}, \end{align*} A B = b − a = ( i + 3 j + 5 k ) − ( 4 i − 9 j − k ) = − 3 i + 12 j + 6 k , and that
D C → = c − d = ( − 2 i − j + 3 k ) − ( i − 13 j − 3 k ) = − 3 i + 12 j + 6 k . \begin{align*} \overrightarrow{DC} &= \mathbf{c} - \mathbf{d} \\ &= \left( -2\mathbf{i} - \mathbf{j} + 3\mathbf{k} \right) - \left( \mathbf{i} - 13 \mathbf{j} - 3 \mathbf{k} \right) \\ &= -3 \mathbf{i} + 12 \mathbf{j} + 6 \mathbf{k}. \end{align*} D C = c − d = ( − 2 i − j + 3 k ) − ( i − 13 j − 3 k ) = − 3 i + 12 j + 6 k .