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Question

A vector perpendicular to the vector (i+2j) and having magnitude 35 units is _____


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Solution

Step 1: Given and assume

The magnitude of r is 35.

Let,

A vector, r=xi+yj perpendicular to A=(i+2j).

Step 2: To find

We have to find a vector perpendicular to the vector (i+2j) and have magnitude 35 units.

Step 3: Calculation

The dot product of r and A must be zero because Cos90°=0.

Example., r.A=(xi+yj).(i+2j)

then,

x+2y=0...(1)

According to the given information,

r=x2+y2=35

By taking square both sides, we get

x2+y2=9×5x2+y2=45

From equation (1),

(-2y)2+y2=454y2+y2=455y2=45y2=9y=3

By substituting the value of “y” in the equation (1), we get

x+2(3)=0x+6=0x=-6

Therefore, the vector perpendicular is r=(6i-3j).


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