$\bar{a}$, $\bar{b}$, $\bar{c}$ are nonzero vectors such that $\bar{a}$ is perpendicular to $\bar{b}$ and $\bar{c}$, $|\bar{a}|=1$, $|\bar{b}|=2$, $|\bar{c}|=1$ and $\bar{b} \cdot \bar{c}=1$. There is nonzero vector $\bar{d}$ coplanar with $\bar{a}+\bar{b}$ and $2\bar{b}-\bar{c}$. If $\bar{d}…

Question

$\bar{a}$, $\bar{b}$, $\bar{c}$ are nonzero vectors such that $\bar{a}$ is perpendicular to $\bar{b}$ and $\bar{c}$, $|\bar{a}|=1$, $|\bar{b}|=2$, $|\bar{c}|=1$ and $\bar{b} \cdot \bar{c}=1$. There is nonzero vector $\bar{d}$ coplanar with $\bar{a}+\bar{b}$ and $2\bar{b}-\bar{c}$. If $\bar{d} \cdot \bar{a}=1$, then $|\bar{d}|^2=$ (Note that $x$ and $y$ are parameters involved when we write $\bar{d}=x(\bar{a}+\bar{b})+y(2\bar{b}-\bar{c})$

Aicharya · Accepted Answer

Given vectors $\bar{a}$, $\bar{b}$, $\bar{c}$ with $\bar{a} \perp \bar{b}$, $\bar{a} \perp \bar{c}$, and known magnitudes $|\bar{a}| = 1$, $|\bar{b}| = 2$, $|\bar{c}| = 1$ with $\bar{b} \cdot \bar{c} = 1$.

The nonzero vector $\bar{d}$ lies in the plane spanned by $\bar{a} + \bar{b}$ and $2\bar{b} - \bar{c}$, so $\bar{d} = x(\bar{a} + \bar{b}) + y(2\bar{b} - \bar{c})$ for scalars $x$, $y$ not both zero.

Expanding gives $\bar{d} = x\bar{a} + (x + 2y)\bar{b} - y\bar{c}$.

Since $\bar{d} \cdot \bar{a} = 1$, and using the orthogonality conditions $\bar{a} \cdot \bar{b} = 0$, $\bar{a} \cdot \bar{c} = 0$, $|\bar{a}|^2 = 1$, we obtain:

$\bar{d} \cdot \bar{a} = x|\bar{a}|^2 = x = 1$.

Substituting $x = 1$ gives $\bar{d} = \bar{a} + (1 + 2y)\bar{b} - y\bar{c}$.

The squared magnitude is:

$|\bar{d}|^2 = (\bar{a} + (1 + 2y)\bar{b} - y\bar{c}) \cdot (\bar{a} + (1 + 2y)\bar{b} - y\bar{c})$.

Expanding and using the given dot products yields:

$|\bar{d}|^2 = 1 + 4(1 + 2y)^2 + y^2 - 2y(1 + 2y) = 13y^2 + 14y + 5$.

This expression matches option A.

$\bar{a}$, $\bar{b}$, $\bar{c}$ are nonzero vectors such that $\bar{a}$ is perpendicular to $\bar{b}$ and…

Solution