What is the vector joining the points (3,1,14) and
(-2,-1,-6)?
Answers
Answer:
please don't keep physics questions because I am from India
Answer:
(
1
√
33
)
⎛
⎜
⎝
−
5
−
2
−
2
⎞
⎟
⎠
Explanation:
If point P is (2,4,4) and Q is (-3,2,2), the vector PQ would be (-5,-2,-2). To find the unit vector, divide vector PQ by its magnitude.
∣
∣
∣
∣
∣
∣
−−→
P
Q
∣
∣
∣
∣
∣
∣
would be
√
(
−
5
)
2
+
(
−
2
)
2
+
(
−
2
)
2
)
=
√
33
. Hence unit vector would be
(
1
√
33
)
⎛
⎜
⎝
−
5
−
2
−
2
⎞
⎟
⎠
Answer link
Douglas K.
Nov 28, 2016
Please see the explanation.
Explanation:
Given the points
(
x
0
,
y
0
,
z
0
)
and
(
x
1
,
y
1
,
z
1
)
You make a vector in the direction from one point to another by subtracting each starting coordinate from its respective ending coordinate:
¯¯¯
V
=
(
x
1
−
x
0
)
ˆ
i
+
(
y
1
−
y
0
)
ˆ
j
+
(
z
1
−
z
0
)
ˆ
k
To make it a unit vector,
ˆ
V
, you divide each component of the vector by the magnitude,
∣
∣
¯¯¯
V
∣
∣
:
∣
∣
¯¯¯
V
∣
∣
=
√
(
x
1
−
x
0
)
2
+
(
y
1
−
y
0
)
2
+
(
z
1
−
z
0
)
2
Therefore the general form for a unit vector is:
ˆ
V
=
x
1
−
x
0
√
(
x
1
−
x
0
)
2
+
(
y
1
−
y
0
)
2
+
(
z
1
−
z
0
)
2
ˆ
i
+
y
1
−
y
0
√
(
x
1
−
x
0
)
2
+
(
y
1
−
y
0
)
2
+
(
z
1
−
z
0
)
2
ˆ
j
+
z
1
−
z
0
√
(
x
1
−
x
0
)
2
+
(
y
1
−
y
0
)
2
+
(
z
1
−
z
0
)
2
ˆ
k
For the given points
(
2
,
4
,
4
)
and
(
−
3
,
2
,
2
)
Compute the magnitude
∣
∣
¯¯¯
V
∣
∣
=
√
(
x
1
−
x
0
)
2
+
(
y
1
−
y
0
)
2
+
(
z
1
−
z
0
)
2
=
√
(
−
3
−
2
)
2
+
(
2
−
4
)
2
+
(
2
−
4
)
2
=
√
33
We need to divide which it the same as multiplying by the reciprocal,
√
33
33
Substitute into the unit vector general form:
ˆ
V
=
(
−
3
−
2
)
√
33
33
ˆ
i
+
(
2
−
4
)
√
33
33
ˆ
j
+
(
2
−
4
)
√
33
33
ˆ
k
Simplify:
ˆ
V
=
−
5
√
33
33
ˆ
i
−
2
√
33
33
ˆ
j
−
2
√
33
33
ˆ
k
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