Question

Let $b$ and $c$ be non-collinear vectors. If $a$ is a vector such that $a\cdot (b+c)=4$ and $a\times (b\times c)=(x^2-2x+6)b+\sin y\cdot c$, then $(x,y)$ lies on the line.

• A. $x+y=0$
• B. $x-y=0$
• C. $x=1$
• D. $y=\pi$

Answer 1

The correct option is C $x=1$

According to the definition of vector triple product.

$\overrightarrow{a}\times \left(\overrightarrow{b}\times \overrightarrow{c}\right)=\left(\overrightarrow{a}.\overrightarrow{c}\right)\overrightarrow{b}-\left(\overrightarrow{a}.\overrightarrow{b}\right)\overrightarrow{c}$

and according to question

$\left(\overrightarrow{a}.\overrightarrow{c}\right)\overrightarrow{b}-\left(\overrightarrow{a}.\overrightarrow{b}\right)\overrightarrow{c}=\left(x^2-2x+6\right)\overrightarrow{b}+\left(siny\right)\overrightarrow{c}$

Now, on compairing

$\overrightarrow{a}.\overrightarrow{c}=x^2-2x+6$

$\overrightarrow{a}.\overrightarrow{b}=-siny$

$\overrightarrow{a}.\left(\overrightarrow{b}+\overrightarrow{c}\right)=4$

$\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{a}.\overrightarrow{c}=4$

$-\sin y+x^2-2x+6=4$

$(x-1{)}^2+1=\sin y$

Minimum value L.H.S. occurs when square term is $0$

thus $x-1=0$ that is $x=1$ and this minimum value will be $(1-1{)}^2+1=1$

and the maximum value of $\sin y$ is $1$.

Thus,

$(x-1{)}^2+1=\sin y$ is possible when

$x=1$ and $y=\frac{\pi }{2}$

$(x,y)=(1,\frac{\pi }{2})$

There will be infinite lines on which $(x,y)=(1,\frac{\pi }{2})$ will lie and $x=1$ is one of them.