Part i is

Part ii, we want to get rid of the to get on its own. To do this we multiply both sides (on the left) by :

.

is given in part iii. and so

Fix the constant .Multiply the matrix by the matrix to get a matrix. Multiply (all entries) by .

This is the vector

Regards,

J.P.

Just to clarify, sorry I know you explained already, could you show me how to answer part c.

]]>Let us note that .

We know that is of magnitude 260 and in the direction of .

, and so .

If we cannot see that 260 is twice 130, or if these magnitudes are not comparable, instead we find a unit vector in the direction of :

,

and

.

Regards,

J.P.

Sorry to disturb you so late, I’m a little confused here on Q 1 (b) i. from the Winter 2017 paper.

How does

give

.

Clearly each coefficient of is multipiled by two, but here they are multiplied by 260?

Regards.

]]>I find

and

.

.

Regards,

J.P.

Just regards Autumn 2018, Q. 1 ii? It says to show the vectors are perpendicular but their dot product does not equal zero?

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