Question

What is the unit vector that is normal to the plane containing $(2i - 3j + k)$ and $(2i + j - 3k)$ ?

Answer 1

Hint : A vector can be expressed as the unit vector when it is divided by its vector magnitude. The unit vector is also known as the normalized vector. Here we will first find the magnitude of the given vector and then find the values for the unit vector and then place the values in the standard formula for the resultant value.

Complete Step By Step Answer:
Now, normal the unit vector can be expressed as:
$\widehat n = \dfrac{{\overline a \times \overline b }}{{\left| {\overline a \times \overline b } \right|}}$ …. (A)
Let $\overline a = (2i - 3j + k)$ and
$\overline b = (2i + j - 3k)$
Now, $$
\overline a \times \overline b = \left| {\begin{array}{*{20}{c}} i&j;&k; \\\ 2&{ - 3}&1 \\\ 2&1&{ - 3} \end{array}} \right|
Expand the above expression –
$\overline a \times \overline b = i(9 - 1) - j( - 6 + 6) + k(2 + 6)$
Terms with the same values and opposite signs cancels each other.
$\overline a \times \overline b = 8i + 8k$ …. (B)
Find the magnitude of the above vector.
$\left| {\overline a \times \overline b } \right| = \sqrt {{{(8)}^2} + {{(8)}^2}}$
Simplify the above equation.
$\left| {\overline a \times \overline b } \right| = \sqrt {64 + 64} \\\ \left| {\overline a \times \overline b } \right| = \sqrt {128} \\\$
Find the square root on the right hand side of the equation. When the same term is multiplied with itself then we get the square of that number. The above expression can be re-written as-
$\left| {\overline a \times \overline b } \right| = \sqrt {{8^2}(2)}$
Square and square root cancel each other on the right hand side of the equation.
$\left| {\overline a \times \overline b } \right| = 8\sqrt 2$ ….. (C)
Now, by using equations (B) and (C),
Equation (A) becomes, $\widehat n = \dfrac{{\overline a \times \overline b }}{{\left| {\overline a \times \overline b } \right|}}$
$\widehat n = \dfrac{{8i + 8k}}{{8\sqrt 2 }}$
Common factors from the numerator and the denominator cancels each other when placed in the form of fraction. Therefore, remove from the numerator and the denominator.
$\widehat n = \dfrac{i}{{\sqrt 2 }} + \dfrac{k}{{\sqrt 2 }}$

Note :
Be good in square and square-root and simplify the resultant value accordingly. Always remember that the square of negative numbers or the positive numbers always give positive value but the magnitude of any vector is always positive.