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Question 1214
Vectors I: Basics, Dot and Cross Products
2014 Paper 1 Question 3 Variant
Question
Generate new
(i)
Find a unit vector
n
{\mathbf{n}}
n
such that
n
×
(
2
i
+
j
−
2
k
)
=
0.
\mathbf{n} \times (2 \mathbf{i} + \mathbf{j} - 2 \mathbf{k}) = \mathbf{0}.
n
×
(
2
i
+
j
−
2
k
)
=
0
.
[2]
(ii)
Find the cosine of the acute angle between
2
i
+
j
−
2
k
{2 \mathbf{i} + \mathbf{j} - 2 \mathbf{k}}
2
i
+
j
−
2
k
and the
y
-
{y\textrm{-}}
y
-
axis.
[1]
Answer
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