Question

Given $A = ai + bj + ck,B = di + 3j + 4k$ and $C = 3i + j - 2k$. If the vectors $A,B$ and $C$ form a triangle such that $A = B + C$ and area $(\Delta ABC) = 5\sqrt {6,}$ then

A. $a = - 8,b = - 4,c = - 2,d = - 11$
B. $a = - 8,b = 4,c = - 2,d = - 11$
C. $a = - 8,b = 4,c = 2,d = - 11$
D. None of the above

Answer 1

Hint- In order to solve this question first we will use the vector sum of two vectors and we will get the value of a, b and c by comparing two vectors further we will evaluate the area of triangle by using the concept as area of triangle formed is equal to $1/2$ times the magnitude of cross product of the $2$ vectors forming the triangle with the vectors being continuous.

Complete step-by-step answer:
Given that
$A = ai + bj + ck \\\ B = di + 3j + 4k \\\$
And $C = 3i + j - 2k$
Here $A,B,C$ are the vectors which represent the sides of the triangle $ABC.$
It is given that $A = B + C$
Now we will substitute the value of $A,B$ and $C$in above equation so we have
$ai + bj + ck = (d + 3)i + 4j + 2k$
By comparing the components of $i$ vector, $j$ vector and $k$ vector , we have
$a = d + 3,b = 4,c = 2$
As we know that the area of the triangle formed is equal to $1/2$ times the magnitude of the cross product of the $2$ vectors forming the triangle with the vectors being continuous.
So first we will evaluate the cross product of any two given vectors as $B{\text{ }} \times {\text{ }}C$
$\Rightarrow$ \left| {\begin{array}{*{20}{c}} i&j;&k; \\\ d&3&4 \\\ 3&1&{ - 2} \end{array}} \right| = - 10i + (2d + 12)j + (d - 9)k
As we know that magnitude of a vector $x{\text{ }} = {\text{ }}pi{\text{ }} + {\text{ }}qj{\text{ }} + {\text{ }}rk$ is given as

$\left| x \right| = \sqrt {{p^2} + {q^2} + {r^2}}$

By applying above formula we have

\sqrt {100 + {{(2d + 12)}^2} + {{(d - 9)}^2}} $$$$\left| {B \times C} \right| = \sqrt {100 + {{(2d + 12)}^2} + {{(d - 9)}^2}}

Therefore the magnitude of cross vector of $B{\text{ }} \times {\text{ }}C$ is $\sqrt {100 + {{(2d + 12)}^2} + {{(d - 9)}^2}}$
Now we will evaluate the value of area of triangle , so from above property we will write the formula as
Area of \Delta ABC{\text{ }} = $$$$\dfrac{1}{2}\left| {B \times C} \right|
Substitute the value of magnitude of vector $B{\text{ }} \times {\text{ }}C$ in above formula we have
$\dfrac{1}{2}\sqrt {\left[ {100 + {{(2d + 12)}^2} + {{(d - 9)}^2}} \right]}$
Area of triangle $ABC.$ is given as
$\Delta ABC{\text{ }} =$ $5\sqrt 6$
So equation becomes as
$\dfrac{1}{2}\sqrt {\left[ {100 + {{(2d + 12)}^2} + {{(d - 9)}^2}} \right]} = 5\sqrt 6$
$\Rightarrow \sqrt {5{d^2} + 30d + 325} = 10\sqrt 6$
Further by taking square of both the sides we have
$5{d^2} + 30d + 325 = 600$
Now by solving quadratic equation we get two solution of $d$ as

$$5{d^2} + 30d - 275 = 0 \\\ \Rightarrow {d^2} + 6d - 55 = 0 \\\$$ $$\Rightarrow (d + 11)(d - 5) = 0 \\\ \Rightarrow d = 5, - 11 \\\$$

By considering value of $d{\text{ }} = {\text{ }}5$
Because
$a = d + 3$
Put the value of $d$ in above equation
So

$$a{\text{ }} = {\text{ }}5 + 3 \\\ a = 8 \\\$$

Now by taking value of $d$ as $- 11$
$a = d + 3$
Put the value of $d$ in above equation
So

$$a = - 11 + 3 \\\ a = - 8 \\\$$

Therefore if $d{\text{ }} = {\text{ }}5$ then $a{\text{ }} = {\text{ }}8{\text{ }},{\text{ }}b{\text{ }} = {\text{ }}4\;$and $c{\text{ }} = {\text{ }}2.$
Or if $d{\text{ }} = {\text{ }} - 11$ then $a{\text{ }} = {\text{ }} - 8{\text{ }},{\text{ }}b{\text{ }} = {\text{ }}4$ and $c{\text{ }} = {\text{ }}2.$
Hence the correct answer is option C.

Note- In algebra, a quadratic equation is any equation that can be rearranged in standard form as $a{x^2} + bx + c = 0$ where $x$ represents an unknown, and $a,{\text{ }}b,{\text{ }}and{\text{ }}c$ represent known numbers, where $a \ne 0$. If$a = 0$, then the equation is linear, not quadratic, as there is no term.