Question

Let \[\vec u = \hat i + \hat j,\vec v = \hat i - \hat j,\] and \[\vec w = \hat i + 2\hat j + 3\hat k\]. If \[\hat n\] is the unit vector such that \[\vec u \cdot \hat n = 0\] and \[\vec v \cdot \hat n = 0\] then \[\left| {\vec w \cdot \hat n} \right|\] is equal to
A. \[3\]
B. \[5\]
C. \[6\]
D. \[10\]

Answer 1

Hint: In the given question, we need to find the value of \[\left| {\vec w \cdot \hat n} \right|\]. For this, we will use the formula of unit vector. Also, for this, we will determine the cross product and modulus of a vector to get the desired result.

Formula used: The following formula used for solving the given question.
The formula of unit vector is given by, \[\hat n = \dfrac{{\vec n}}{{\left| {\vec n} \right|}}\]
Also, if \[\vec u = a\hat i + b\hat j + c\hat k,\vec v = p\hat i + q\hat j + r\hat k\] then \[\vec u \times \vec v = \left| {\begin{array}{*{20}{c}}i&j&k\\a&b&c\\p&q&r\end{array}} \right|\]
That is \[\vec u \times \vec v = \left| {\begin{array}{*{20}{c}}i&j&k\\a&b&c\\p&q&r\end{array}} \right| = i\left( {br - qc} \right) - j\left( {ar - pc} \right) + k\left( {aq - pb} \right)\]
Also, the modulus of a vector \[\vec u = a\hat i + b\hat j + c\hat k\]is given by
\[\left| {\vec u} \right| = \sqrt {{a^2} + {b^2} + {c^2}} \]

Complete step by step solution: We know that \[\vec u = \hat i + \hat j,\vec v = \hat i - \hat j,\]and \[\vec w = \hat i + 2\hat j + 3\hat k\]
Also, we have \[\vec u \cdot \hat n = 0\] and \[\vec v \cdot \hat n = 0\]
Thus, we can say that \[\hat n = \dfrac{{\vec u \times \vec v}}{{\left| {\vec u \times \vec v} \right|}}\]
Here, \[\hat n\] is the unit vector.
Now, \[\vec u \times \vec v = \left| {\begin{array}{*{20}{c}}i&j&k\\1&1&0\\1&{ - 1}&0\end{array}} \right|\]
By simplifying, we get
\[\vec u \times \vec v = i\left( 0 \right) - j\left( 0 \right) + k\left( { - 1 - 1} \right)\]
By simplifying, we get
\[\vec u \times \vec v = - 2k\]
Now, we will find the modulus of a vector \[\vec u \times \vec v = - 2k\]
So, we get
\[\left| {\vec u \times \vec v} \right| = \sqrt {{{\left( 2 \right)}^2}} \]
By simplifying, we get
\[\left| {\vec u \times \vec v} \right| = 2\]
Thus, \[\hat n = \dfrac{{ - 2k}}{2}\]
\[\hat n = - k\]
Now, we will find the value of \[\left| {\vec w \cdot \hat n} \right|\]
\[\left| {\vec w \cdot \hat n} \right| = \left| {\dfrac{{\left( {\hat i + 2\hat j + 3\hat k} \right) \cdot \left( { - k} \right)}}{2}} \right|\]
\[\left| {\vec w \cdot \hat n} \right| = \left| {\dfrac{{\left( { - 6} \right)}}{2}} \right|\]
By simplifying, we get
\[\left| {\vec w \cdot \hat n} \right| = 3\]
Hence, if \[\vec u = \hat i + \hat j,\vec v = \hat i - \hat j,\]and \[\vec w = \hat i + 2\hat j + 3\hat k\] and \[\hat n\] is the unit vector such that \[\vec u \cdot \hat n = 0\] and \[\vec v \cdot \hat n = 0\] then \[\left| {\vec w \cdot \hat n} \right|\] is equal to \[3\].

Thus, Option (A) is correct.

Additional Information: The definition of a vector is an entity with both magnitude and direction. The movement of an object between two points is described by a vector. The directed line segment can be used to geometrically represent vector mathematics. The magnitude of a vector is the length of the directed line segment, and the vector's direction is indicated by the angle at which it is inclined. A vector's "Tail" (the point where it begins) and "Head" (the point where it ends and has an arrow) are its respective names.

Note: Many students make mistakes in the simplification part as well as writing the formula of unit vectors. This is the only way through which we can solve the example in the simplest way. Also, it is necessary to find the appropriate value of \[\left| {\vec u \times \vec v} \right|\]to get the value of \[\left| {\vec w \cdot \hat n} \right|\].