Question

Let \[\overrightarrow a = \widehat i - \widehat j\], \[\overrightarrow b = \widehat i + \widehat j + \widehat k\] and \[\overrightarrow c \] be a vector such that \[\overrightarrow a \times \overrightarrow c + \overrightarrow b = 0\] and \[\overrightarrow a \cdot \overrightarrow c = 4\], then find\[{\left| {\overrightarrow c } \right|^2} \]

Answer 1

Hint: From the question, we can see some hints where we can use some simple identities and formula which are used in vector properties such that vector and scalar products and the result will be found by using these identities by putting the known values that are given in the question. Suppose if we headed with substitution method, like taking unknow vector as \[\overrightarrow c = x\widehat i + y\widehat j + z\widehat k\] should lead us complication path. So, just think some tricks by use the given conditions which is lead us through the clarity way.


Formula used: Since vector algebra is a very interesting subject we can solve the problems in this chapter very easily. We have to just know some tricks and identities, formulas to solve these types of problems. In the vector product, \[\overrightarrow a \times \overrightarrow b = - \overrightarrow b \times \overrightarrow a \] and in the scalar product, \[\overrightarrow a \cdot \overrightarrow b = \overrightarrow b \cdot \overrightarrow a \]
Next, \[\left( {\overrightarrow a \times \overrightarrow b } \right) \times \overrightarrow c = \left( {\overrightarrow a \cdot \overrightarrow c } \right)\overrightarrow b - \left( {\overrightarrow b \cdot \overrightarrow c } \right)\overrightarrow a \] and the magnitude of a vector is modulus of the vector.
For example, consider a vector \[\overrightarrow p = x\widehat i + y\widehat j + z\widehat k\] and its magnitude will be defined as
\[\left| {\overrightarrow p } \right| = \sqrt {{x^2} + {y^2} + {z^2}} \]

Complete step by step solution: Since vector algebra is a very interesting subject we can solve the problems in this chapter very easily. We have to just know some tricks and identities, formulas to solve these types of problems. In the vector product, \[\overrightarrow a \times \overrightarrow b = - \overrightarrow b \times \overrightarrow a \] and in the scalar product, \[\overrightarrow a \cdot \overrightarrow b = \overrightarrow b \cdot \overrightarrow a \]
Next, \[\left( {\overrightarrow a \times \overrightarrow b } \right) \times \overrightarrow c = \left( {\overrightarrow a \cdot \overrightarrow c } \right)\overrightarrow b - \left( {\overrightarrow b \cdot \overrightarrow c } \right)\overrightarrow a \] and the magnitude of a vector is modulus of the vector.
For example, consider a vector \[\overrightarrow p = x\widehat i + y\widehat j + z\widehat k\] and its magnitude will be defined as
\[\left| {\overrightarrow p } \right| = \sqrt {{x^2} + {y^2} + {z^2}} \]
Complete step by step answer:
Given that \[\overrightarrow a \times \overrightarrow c + \overrightarrow b = 0\] and \[\overrightarrow a \cdot \overrightarrow c = 4\]
First take \[\overrightarrow a \times \overrightarrow c + \overrightarrow b = 0\]. From this by using the identity we can write it as,
     \[\begin{array}{c}\overrightarrow a \times \overrightarrow c + \overrightarrow b = \overrightarrow a \times \overrightarrow c = - \overrightarrow b \end{array}\]
Now, multiply by \[\overrightarrow a \] on both sides such that we get,
\[\left( \vec{a}\times \vec{c} \right)\times \vec{a}=-\vec{b}\times \vec{a}\]
\[\left( \vec{a}\cdot \vec{a} \right)\vec{c}-\left( \vec{c}\cdot \vec{a} \right)\vec{a}=\vec{a}\times \vec{b}\]
\[\left( \hat{i}-\hat{j} \right)\cdot \left( \hat{i}-\hat{j} \right)\vec{c}-\left( 4 \right)\left( \hat{i}-\hat{j} \right)=\vec{a}\times \vec{b}\]
\[\left( 1+1 \right)\vec{c}-4\hat{i}+4\hat{j}=\vec{a}\times \vec{b}\]
\[2\vec{c}-4\hat{i}+4\hat{j}=\vec{a}\times \vec{b}\]
Next, let us find \[\overrightarrow a \times \overrightarrow b \]
\[\overrightarrow a \times \overrightarrow b\] = $\begin{vmatrix}{\widehat i}&{\widehat j}&{\widehat k}\\1&{- 1}&0\\1&1&1 \end{vmatrix}$
\[= \widehat i\left( { - 1 - 0} \right) - \widehat j\left( {1 - 0} \right) + \widehat k\left( {1 + 1} \right)\]
\[= - \widehat i - \widehat j + 2\widehat k\]
Now from the equation (1.1), we have
\[2\overrightarrow c - 4\widehat i + 4\widehat j = - \widehat i - \widehat j + 2\widehat k\]
\[2\overrightarrow c = - \widehat i - \widehat j + 2\widehat k + 4\widehat i - 4\widehat j\]
\[2\overrightarrow c = 3\widehat i - 5\widehat j + 2\widehat k\]
\[\overrightarrow c = \dfrac{3}{2}\widehat i - \dfrac{5}{2}\widehat j + \dfrac{2}{2}\widehat k\]
This implies that,
\[\left| {\overrightarrow c } \right| = \sqrt {{{\left( {\dfrac{3}{2}} \right)}^2} + {{\left( { - \dfrac{5}{2}} \right)}^2} + {{\left( 1 \right)}^2}}\]
\[= \sqrt {\dfrac{9}{4} + \dfrac{{25}}{4} + 1}\]
\[= \sqrt {\dfrac{{34}}{4} + 1}\]
\[= \sqrt {\dfrac{{38}}{4}}\]
\[\left| {\overrightarrow c } \right| = \sqrt {\dfrac{{19}}{2}}\]
Next, square on both sides:
\[{\left| {\overrightarrow c } \right|^2} = \dfrac{{19}}{2}\]
\[{\left| {\overrightarrow c } \right|^2} = 9.5\]
This is the required answer.


Note: We should don’t be confused about using the identities and properties of vector and should be careful in the signs of the terms as there may be a missing of plus and minus and calculations. So, just avoid the errors.