For the vectors a = (3, 12) and b = (6, 9), find orth ab.

Solution:

Given, vectors a = (3, 12) and b = (6, 9)

We have to find orth ab.

Orthₐb = b - projₐb

projₐb = \(\frac{a.b}{\left | a \right |^{2}}*a\)

Now, a.b = (3, 12).(6, 9)

= 3(6) + 12(9)

= 18 + 108

a.b = 126

|a| is the modulus of vector a.

\(\left | a \right |=\sqrt{3^{2}+12^{2}}\)

\(\left | a \right |=\sqrt{9+144}\)

\(\left | a \right |=\sqrt{153}\)

\(\left | a \right |^{2}=(\sqrt{153})^{2}\)

\(\left | a \right |^{2}=153\)

projₐb = \(\frac{126}{153}.(3, 12)\)

= \(\frac{126\times 3}{153},\frac{126\times 12}{153}\)

= \(\frac{126}{51},\frac{126\times 4}{51}\)

projₐb = \(\frac{126}{51},\frac{504}{51}\)

Orthₐb = (3, 12) - \(\frac{126}{51},\frac{504}{51}\)

\(= 3-\frac{126}{51},12-\frac{504}{51}\)

= \(\frac{153 - 126}{51},\frac{612 - 504}{51}\)

= \(\frac{27}{51},\frac{108}{51}\)

Orthₐb = (9/17, 36/17)

Therefore, Orthₐb = (9/17, 36/17)

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For the vectors a = (3, 12) and b = (6, 9), find orth ab.

Summary:

For the vectors a = (3, 12) and b = (6, 9), orth ab is (9/17, 36/17).