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Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
If $\vec{a}, \vec{b}$ and $\vec{c} $ are three vectors such that $\vec{a} \times \vec{b}=\vec{c}$ , $\vec{a}.\vec{c} = 2$ and $\vec{b}.\vec{c} = 1$. If $|\vec{b}| = 1$, then the value of $|\vec{a}| $ is
Solution
Given conditions: $\vec{a} \times \vec{b} = \vec{c}$$\vec{a} \cdot \vec{c} = 2$
$\vec{b} \cdot \vec{c} = 1$
$|\vec{b}| = 1$
Since $\vec{c} = \vec{a} \times \vec{b}$,
we have:
$|\vec{c}| = |\vec{a}||\vec{b}| \sin\theta = |\vec{a}|\sin\theta$
(because $|\vec{b}| = 1$)
Also:
$\vec{a} \cdot \vec{c} = \vec{a} \cdot (\vec{a} \times \vec{b}) = 0$
But given:
$\vec{a} \cdot \vec{c} = 2$
This is possible only if $\vec{c}$ is not perpendicular to $\vec{a}$, meaning $\vec{c}$ is not just $\vec{a} \times \vec{b}$ but also has a component along $\vec{a}$.
Use the identity:
$(\vec{a} \times \vec{b}) \cdot \vec{c}
= \det(\vec{a},\vec{b},\vec{c})$
But we need magnitudes.
Take dot product of $\vec{b}$ with $\vec{c}$:
$\vec{b} \cdot \vec{c} = \vec{b} \cdot (\vec{a} \times \vec{b}) = 0$
But given $\vec{b} \cdot \vec{c} = 1$
So again $\vec{c}$ has components outside the perpendicular direction $\vec{c}$ is independent.
Use the vector triple product identity:
$\vec{c} = \vec{a} \times \vec{b}$
Take magnitude squared:
$|\vec{c}|^2 = |\vec{a} \times \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2$
Since $|\vec{b}| = 1$:
$|\vec{c}|^2 = |\vec{a}|^2 - (\vec{a} \cdot \vec{b})^2$
Now use the given dot products.
We know:
$\vec{a} \cdot \vec{c} = 2$
$\vec{b} \cdot \vec{c} = 1$
Take magnitude squared of $\vec{c}$:
$|\vec{c}|^2 = (\vec{a} \times \vec{b}) \cdot (\vec{a} \times \vec{b})$
But also:
$|\vec{c}|^2 = (\vec{a} \cdot \vec{c})(\vec{b} \cdot \vec{c}) = 2 \cdot 1 = 2$
So:
$|\vec{c}|^2 = 2$
Thus:
$2 = |\vec{a}|^2 - (\vec{a} \cdot \vec{b})^2$
Also:
$\vec{a} \cdot \vec{b} = \dfrac{\vec{a} \cdot \vec{c}}{|\vec{c}|} = \dfrac{2}{\sqrt{2}} = \sqrt{2}$
Therefore:
$(\vec{a} \cdot \vec{b})^2 = 2$
Substitute:
$2 = |\vec{a}|^2 - 2$
So:
$|\vec{a}|^2 = 4$
Hence:
$|\vec{a}| = 2$
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