Question

If ${\left| {\overrightarrow C } \right|^2} = 60$and $\overrightarrow C \times (\hat i + 2\hat j + 5\hat k) = 0$, then the value of $\overrightarrow C \bullet ( - 7\hat i + 2\hat j + 3\hat k)$is
A. $4\sqrt 2$
B. $24$
C. $12\sqrt 2$
D. $12$

Answer 1

We assume the vector $\overrightarrow C = a\hat i + b\hat j + c\hat k$and use property that dot product of the vector to itself gives the square of its magnitude, which gives us an equation in a,b and c. Then we use the method of cross product of two vectors to solve $\overrightarrow C \times (\hat i + 2\hat j + 5\hat k) = 0$which will give us other three equations in a,b and c. Using these equations we find the values of constants and write the vector C.

• $\overrightarrow a \bullet \overrightarrow a = {\left| {\overrightarrow a } \right|^2}$
• Dot product of two vectors $({a_1}\hat i + {b_1}\hat j + {c_1}\hat k),({a_2}\hat i + {b_2}\hat j + {c_2}\hat k) = {a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}$
• Cross product of two vectors $({a_1}\hat i + {b_1}\hat j + {c_1}\hat k),({a_2}\hat i + {b_2}\hat j + {c_2}\hat k)$is given by writing the terms in form of a determinant \left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k} \\\ {{a_1}}&{{b_1}}&{{c_1}} \\\ {{a_2}}&{{b_2}}&{{c_2}} \end{array}} \right| = ({b_1}{c_2} - {b_2}{c_1})\hat i - ({a_1}{c_2} - {a_2}{c_1})\hat j + ({a_1}{b_2} - {a_2}{b_1})\hat k.

Complete step-by-step answer:
Let us assume the vector $\overrightarrow C = a\hat i + b\hat j + c\hat k$
No we know that the dot product of vector to itself gives the square of magnitude of the vector, i.e. $\overrightarrow a \bullet \overrightarrow a = {\left| {\overrightarrow a } \right|^2}$. So we take the dot product of the vector $\overrightarrow C = a\hat i + b\hat j + c\hat k$ with itself.
$\Rightarrow \overrightarrow C \bullet \overrightarrow C = {\left| {\overrightarrow C } \right|^2}$
Substituting the value of $\overrightarrow C = a\hat i + b\hat j + c\hat k$ and ${\left| {\overrightarrow C } \right|^2} = 60$
$\Rightarrow (a\hat i + b\hat j + c\hat k) \bullet (a\hat i + b\hat j + c\hat k) = 60$
Using the formula for dot product of two vectors $({a_1}\hat i + {b_1}\hat j + {c_1}\hat k),({a_2}\hat i + {b_2}\hat j + {c_2}\hat k) = {a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}$
We can write
$\Rightarrow a \times a + b \times b + c \times c = 60$
$\Rightarrow {a^2} + {b^2} + {c^2} = 60$ … (1)
Now we take the cross product of $\overrightarrow C \times (\hat i + 2\hat j + 5\hat k) = 0$, where $\overrightarrow C = a\hat i + b\hat j + c\hat k$.
$\Rightarrow (a\hat i + b\hat j + c\hat k) \times (\hat i + 2\hat j + 5\hat k) = 0$
We can write the cross product in determinant form as

{\hat i}&{\hat j}&{\hat k} \\\ a&b;&c; \\\ 1&2&5 \end{array}} \right| = (5b - 2c)\hat i - (5a - c)\hat j + (2a - b)\hat k$$ Since the cross product is equal to zero, $$ \Rightarrow (5b - 2c)\hat i - (5a - c)\hat j + (2a - b)\hat k = 0\hat i + 0\hat j + 0\hat k$$ Two vectors are equal if their constant terms are equal $$ \Rightarrow 5b - 2c = 0,5a - c = 0,2a - b = 0$$ Solving each equation separately Firstly $$5b - 2c = 0$$ Shifting one value to other side of the equation we can write $$ \Rightarrow 5b = 2c$$ Squaring both sides of the equation

\Rightarrow {(5b)^2} = {(2c)^2} \\
\Rightarrow 25{b^2} = 4{c^2} \\

Secondly $$5a - c = 0$$ Shifting one value to other side of the equation we can write $$ \Rightarrow 5a = c$$ Squaring both sides of the equation

\Rightarrow {(5a)^2} = {(c)^2} \\
\Rightarrow 25{a^2} = {c^2} \\

Thirdly $$2a - b = 0$$ Shifting one value to other side of the equation we can write $$ \Rightarrow 2a = b$$ Squaring both sides of the equation

\Rightarrow {(2a)^2} = {(b)^2} \\
\Rightarrow 4{a^2} = {b^2} \\

Now substitute the value of $${b^2},{c^2}$$from equations (3) and (4) in equation (1)

\Rightarrow {a^2} + 4{a^2} + 25{a^2} = 60 \\
\Rightarrow 30{a^2} = 60 \\

$$Divide both sides by 30$$

\Rightarrow \dfrac{{30{a^2}}}{{30}} = \dfrac{{60}}{{30}} \\
\Rightarrow {a^2} = 2 \\

$$a = \sqrt 2 $$ Now substitute the value of $${a^2} = 2$$in equation (3) and (4) From equation (3): $${c^2} = 25 \times 2 = 50$$ $$c = \sqrt {50} = \sqrt {{5^2} \times 2} = 5\sqrt 2 $$ From equation (4): $${b^2} = 4 \times 2 = 8$$ $$b = \sqrt 8 = \sqrt {{2^2} \times 2} = 2\sqrt 2 $$ So, the vector becomes $$\overrightarrow C = \sqrt 2 \hat i + 2\sqrt 2 \hat j + 5\sqrt 2 \hat k$$ Now we calculate the dot product of $$\overrightarrow C \bullet ( - 7\hat i + 2\hat j + 3\hat k)$$ is $$ \Rightarrow (\sqrt 2 \hat i + 2\sqrt 2 \hat j + 5\sqrt 2 \hat k) \bullet ( - 7\hat i + 2\hat j + 3\hat k) = \sqrt 2 \times ( - 7) + 2\sqrt 2 \times (2) + 5\sqrt 2 \times (3)$$ Multiplying the terms on RHS $$ \Rightarrow (\sqrt 2 \hat i + 2\sqrt 2 \hat j + 5\sqrt 2 \hat k) \bullet ( - 7\hat i + 2\hat j + 3\hat k) = - 7\sqrt 2 + 4\sqrt 2 + 15\sqrt 2 $$ Take $$\sqrt 2 $$common in RHS of the equation

\Rightarrow (\sqrt 2 \hat i + 2\sqrt 2 \hat j + 5\sqrt 2 \hat k) \bullet ( - 7\hat i + 2\hat j + 3\hat k) = \sqrt 2 ( - 7 + 4 + 15) \\
\Rightarrow (\sqrt 2 \hat i + 2\sqrt 2 \hat j + 5\sqrt 2 \hat k) \bullet ( - 7\hat i + 2\hat j + 3\hat k) = 12\sqrt 2 \\

**So, the correct option is C.** **Note:** Students make mistake of considering dot product as normal multiplication and they multiply all the values from first bracket to all values in second bracket one by one which is wrong because when we take dot product of two vectors, we only write the multiplication of numbers in same direction because $$\hat i.\hat i = \hat j.\hat j = \hat k.\hat k = 1$$, as multiplication of numbers in different direction will give us zero because $$\hat i.\hat j = \hat j.\hat k = \hat k.\hat i = 0$$. Students are likely to make mistakes calculating the cross product as the negative sign before the $$\hat j$$is already existing and some students write another negative sign when solving the determinant which makes it positive and then the answer comes out to be wrong.