Let a=2i-3j+k,, b=3i+2j+5k and a vector c be such that (a-c) b=-18i-3j+12k and a c=3.

Question

Let $\overset{	o }{a}=2\hat{i}-3\hat{j}+\hat{k},\, \overset{	o \leftarrow }{b}=3\hat{i}+2\hat{j}+5\hat{k}$ and a vector $\overset{	o }{c}$ be such that $(\vec{a}-\vec{c})	imes \vec{b}=-18\hat{i}-3\hat{j}+12\hat{k}$ and $\vec{a}\cdot \vec{c}=3$. If $\vec{b}	imes \vec{c}=\vec{d}$, then $|\vec{a}\cdot \vec{d}|$ is equal to :

Accepted Answer

To solve the problem, we start with the given equation:

$(\vec{a}-\vec{c})	imes \vec{b}=-18\hat{i}-3\hat{j}+12\hat{k}.$

Expanding this, we have:

$\vec{a}	imes \vec{b}-\vec{c}	imes \vec{b}=-18\hat{i}-3\hat{j}+12\hat{k}.$

Let $\vec{d}=\vec{c}	imes \vec{b}$. Substituting this into the equation, we get:

$\vec{a}	imes \vec{b}+\vec{d}=-18\hat{i}-3\hat{j}+12\hat{k}.$

Now, we take the dot product of both sides with $\vec{a}$:

$\vec{a}\cdot (\vec{a}	imes \vec{b})+\vec{a}\cdot \vec{d}=\vec{a}\cdot (-18\hat{i}-3\hat{j}+12\hat{k}).$

Since the dot product $\vec{a}\cdot (\vec{a}	imes \vec{b})$ is zero (the dot product of any vector with the cross product of itself with another vector is zero), we have:

$\vec{a}\cdot \vec{d}=-36+9+12=-15.$

Thus, the magnitude of $\vec{a}\cdot \vec{d}$ is:

$|\vec{a}\cdot \vec{d}|=15.$

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