Question

Let the vector $\overrightarrow a = \overrightarrow i - 2\overrightarrow j + \overrightarrow k$and vector $\overrightarrow b = \overrightarrow i - \overrightarrow j + \overrightarrow k$be two vectors. If vector $\overrightarrow c$ is a vector such that $\overrightarrow b \times \overrightarrow c = \overrightarrow b \times \overrightarrow a$ and $\overrightarrow c .\overrightarrow a = 0$ then $\overrightarrow c .\overrightarrow b$ equal to?
$A)\dfrac{{ - 3}}{2}$
$B) - 1$
$C)\dfrac{1}{2}$
$D)\dfrac{{ - 1}}{2}$

Answer 1

First, the vector is the magnitude and direction both values are contained.
Since the cross product of the two vectors are the binary operation in the three-dimensional space, which is defined as $a \times b$ the vector $c$ that is perpendicular to both a and b with the direction of right-hand rule and magnitude equals the area of the parallelogram
The dot product is known as the sum of the products of the corresponding entries of the two vectors.
Formula used: $\left| {\overrightarrow a } \right| = \sqrt {{{(1)}^2} + {{( - 2)}^2} + {{(1)}^2}}$ with the vector values $i,j,k$ where $\overrightarrow a = \overrightarrow i - 2\overrightarrow j + \overrightarrow k$

Complete step by step answer:
From the given that the vector $\overrightarrow a = \overrightarrow i - 2\overrightarrow j + \overrightarrow k$ and vector $\overrightarrow b = \overrightarrow i - \overrightarrow j + \overrightarrow k$ be two vectors.
By the taking modulus of the vectors we get, $\left| {\overrightarrow a } \right| = \sqrt {{{(1)}^2} + {{( - 2)}^2} + {{(1)}^2}} \Rightarrow \sqrt 6$ and $\left| {\overrightarrow b } \right| = \sqrt {{{(1)}^2} + {{( - 1)}^2} + {{(1)}^2}} \Rightarrow \sqrt 3$
Now we are going to act the dot product of the two vectors, which is $\overrightarrow a .\overrightarrow b = (\overrightarrow i - 2\overrightarrow j + \overrightarrow k ).(\overrightarrow i - \overrightarrow j + \overrightarrow k )$ which is the values will be multiplied into the corresponding vectors and add the overall values we get, $\overrightarrow a .\overrightarrow b = (\overrightarrow i - 2\overrightarrow j + \overrightarrow k ).(\overrightarrow i - \overrightarrow j + \overrightarrow k ) \Rightarrow 1 + 2 + 1 \Rightarrow 4$
From the given that $\overrightarrow b \times \overrightarrow c = \overrightarrow b \times \overrightarrow a$, now rewriting this we get, $\overrightarrow b \times \overrightarrow c = \overrightarrow b \times \overrightarrow a \Rightarrow \overrightarrow b \times \overrightarrow c - \overrightarrow b \times \overrightarrow a = 0$
Taking the vector b in common we get, $\overrightarrow b \times \overrightarrow c - \overrightarrow b \times \overrightarrow a = 0 \Rightarrow \overrightarrow b \times (\overrightarrow c - \overrightarrow a ) = 0$ which means the vector b is parallel to the vector c minus a, then we can rewrite this as, $\overrightarrow b \times (\overrightarrow c - \overrightarrow a ) = 0 \Rightarrow c - a = kb$(where k be any constant)
Now applying the an into the right-hand side, we get, $c - a = kb \Rightarrow c = kb + a$
Also, from the given that we have, $\overrightarrow c .\overrightarrow a = 0$ and substitute the value of c in the dot product we get, $(kb + a).\overrightarrow a = 0$now simplifying and equating the constant term in separate thus we get, $k = \dfrac{{ - \left| {{a^2}} \right|}}{{a.b}}$
Applying the values that, $\left| {\overrightarrow a } \right| = \sqrt {{{(1)}^2} + {{( - 2)}^2} + {{(1)}^2}} \Rightarrow \sqrt 6$and $\overrightarrow a .\overrightarrow b = (\overrightarrow i - 2\overrightarrow j + \overrightarrow k ).(\overrightarrow i - \overrightarrow j + \overrightarrow k ) \Rightarrow 1 + 2 + 1 \Rightarrow 4$, we get, $k = \dfrac{{ - \left| {{a^2}} \right|}}{{a.b}} \Rightarrow \dfrac{{ - 6}}{4} = \dfrac{{ - 3}}{2}$
Therefore, we get the value for the constant now substitute that in $c = kb + a \Rightarrow a - \dfrac{3}{2}b$
Appling the vector a and c, we get c. thus $c = a - \dfrac{3}{2}b \Rightarrow \overrightarrow i - 2\overrightarrow j + \overrightarrow k - \dfrac{3}{2}(\overrightarrow i - \overrightarrow j + \overrightarrow k )$
Further soling this we get, $c = \overrightarrow i - 2\overrightarrow j + \overrightarrow k - \dfrac{3}{2}(\overrightarrow i - \overrightarrow j + \overrightarrow k ) \Rightarrow \dfrac{{ - 1}}{2}(\overrightarrow i + \overrightarrow j + \overrightarrow k )$
Finally, the given question asks us to find them $\overrightarrow c .\overrightarrow b$ since we know the vector c and b
Applying we get, $\overrightarrow c .\overrightarrow b = \dfrac{{ - 1}}{2}(\overrightarrow i + \overrightarrow j + \overrightarrow k )(\overrightarrow i - \overrightarrow j + \overrightarrow k )$taking the dot product with the corresponding vector we get, $\overrightarrow c .\overrightarrow b = \dfrac{{ - 1}}{2}(1 - 1 + 1) = \dfrac{{ - 1}}{2}$
Therefore, the dot product value of $\overrightarrow c .\overrightarrow b = \dfrac{{ - 1}}{2}$

So, the correct answer is “Option D”.

Note: while dot using the dot product, we have to multiply the corresponding vectors like is $(\overrightarrow i + \overrightarrow j ).(\overrightarrow i - \overrightarrow j )$
First multiplied the vector i, in that there is one and one are the numbers thus we get one, second multiple the vector j, in that one and minus one are the numbers, multiplying we get minus one,
In the final step add the one and minus one we get zero, this process is mathematical represent as $(\overrightarrow i + \overrightarrow j ).(\overrightarrow i - \overrightarrow j ) = 1.1 + 1.( - 1) \Rightarrow 1 - 1 = 0$