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The vectors $\vec{a}=x \vec{i}+y \vec{j}+z \vec{k}, \vec{b}=\hat{k}, \vec{c}$ are such that they
form a right handed system. What is $\vec{c}$ equal to?
Options:
form a right handed system. What is $\vec{c}$ equal to?
Solution:
1109 Upvotes
Verified Answer
The correct answer is:
$y \hat{\boldsymbol{i}}-x \hat{\boldsymbol{j}}$
We know, scalar triple product $(\vec{a} \times \vec{b}) \cdot \vec{c}$ is positive or
negative according as $\vec{a}, \vec{b}, \vec{c}$ form a right handed or
left handed system respectively. consider option (a)
Let $\vec{c}=\vec{j}$
$\therefore[\vec{a} \vec{b} \vec{c}]=\left|\begin{array}{lll}x & y & z \\ 0 & 0 & 1 \\ 0 & 1 & 0\end{array}\right|=x(-1)-y(0)+z(0)$
$=-x$
option (b)
Let $\vec{c}=y \hat{j}-x \hat{k}$
$\begin{aligned} \therefore \quad[\vec{a} \vec{b} \vec{c}]=\left|\begin{array}{ccc}x & y & z \\ 0 & 0 & 1 \\ 0 & y-x\end{array}\right| &=x(-y)-y(0)+z(0) \\ &=-x y \end{aligned}$
option (c)
Let $\vec{c}=y \hat{i}-x \hat{j}$
$\begin{aligned} \therefore \quad[\vec{a} \vec{b} \vec{c}] &=\left|\begin{array}{rrr}x & y & z \\ 0 & 0 & 1 \\ y & -x & 0\end{array}\right| \\ &=x(x)-y(-y)+z(0)=x^{2}+y^{2} \end{aligned}$
Since, scalar triple product is positive when $\vec{c}=y \hat{i}-x \hat{j}$
$\therefore$ option (c) is correct.
negative according as $\vec{a}, \vec{b}, \vec{c}$ form a right handed or
left handed system respectively. consider option (a)
Let $\vec{c}=\vec{j}$
$\therefore[\vec{a} \vec{b} \vec{c}]=\left|\begin{array}{lll}x & y & z \\ 0 & 0 & 1 \\ 0 & 1 & 0\end{array}\right|=x(-1)-y(0)+z(0)$
$=-x$
option (b)
Let $\vec{c}=y \hat{j}-x \hat{k}$
$\begin{aligned} \therefore \quad[\vec{a} \vec{b} \vec{c}]=\left|\begin{array}{ccc}x & y & z \\ 0 & 0 & 1 \\ 0 & y-x\end{array}\right| &=x(-y)-y(0)+z(0) \\ &=-x y \end{aligned}$
option (c)
Let $\vec{c}=y \hat{i}-x \hat{j}$
$\begin{aligned} \therefore \quad[\vec{a} \vec{b} \vec{c}] &=\left|\begin{array}{rrr}x & y & z \\ 0 & 0 & 1 \\ y & -x & 0\end{array}\right| \\ &=x(x)-y(-y)+z(0)=x^{2}+y^{2} \end{aligned}$
Since, scalar triple product is positive when $\vec{c}=y \hat{i}-x \hat{j}$
$\therefore$ option (c) is correct.
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