Let a = i + 4 j + 2 k, b = 3 i - 2 j + 7 k and c = 2 i - j + 4 k.… - Vidyadip

Question

Let $\vec a = \hat i + 4\hat j + 2\hat k$, $\vec b = 3\hat i - 2\hat j + 7\hat k$ and $\vec c = 2\hat i - \hat j + 4\hat k$. Find a vector $\vec d$ which is perpendicular to both $\vec a$ and $\vec b$, and $\vec c.\vec d = 15$.

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Answer

Given: Vectors $\vec a = \hat i + 4\hat j + 2\hat k$ and $\vec b = 3\hat i - 2\hat j + 7\hat k$
We know that the cross-product of two vectors, $\vec a \times \vec b$ is a vector perpendicular to both $\vec a$ and $\vec b$
Hence, vector $\vec d$ which is also perpendicular to both $\vec a$ and $\vec b$ is $\vec d = \lambda \left( {\vec a \times \vec b} \right)$ where $\lambda = 1$ or some other scalar.
Therefore, $\vec d = \lambda \left| {\begin{array}{*{20}{c}} \vec i&\vec j&\vec k \\ 1&4&2 \\ 3&{ - 2}&7 \end{array}} \right|$
$= \lambda \left[ {\hat i\left( {28 + 4} \right) - \hat j\left( {7 - 6} \right) + \hat k\left( { - 2 - 12} \right)} \right]$
$ \Rightarrow \vec d = 32\lambda \hat i - \lambda \hat j - 14\lambda \hat k$...(i)
Now given $\vec c = 2\hat i - \hat j + 4\hat k$ and $\vec c.\vec d = 15$
$\vec c.\vec d = 15$
$= 2\left( {32\lambda } \right) + \left( { - 1} \right)\left( { - \lambda } \right) + 4\left( { - 14\lambda } \right) = 15$
$ \Rightarrow 64\lambda + \lambda - 56\lambda = 15$
$ \Rightarrow 9\lambda = 15$
$ \Rightarrow \lambda = \frac{{15}}{9}$
$ \Rightarrow \lambda = \frac{5}{3}$
Putting $\lambda = \frac{5}{3}$ in eq. (i), we get
$\vec d = \frac{5}{3}\left[ {32\hat i - \hat j - 14\hat k} \right]$
$\Rightarrow \vec d = \frac{1}{3}\left[ {160\hat i - 5\hat j - 70\hat k} \right]$

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