If either a=0 or b=0, then a =0. Is the converse true? Justify your answer with an example.

Given: Either a=0 or b=0 |a |= |0 |=0 or |b |= |0 |=0 ........(i) |a |= |a | |b | =0 0. =0 [Using eq.(i) ] a =0 [By definition of zero vector] But the converse is not true. Let a=i+j+k |a |=1+1+1=3 0 a is a non-zero vector. Let b=2 (i+j+k )=2i+2j+2k |b |=4+4+4=12=23 0 b is a non-zero vector. But a b= i&j&k 1&1&1 2&2&2 Taking 2 common from R_ 3 =a b= i&j&k 1&1&1 1&1&1 =0 [ _2 and R_ 3 are identical ]

If either a=0 or b=0, then a =0. Is the converse true? Justify yo…

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Question

If either $\vec{a}=\vec{0}\ \ \text{or}\ \ \vec{b}=\vec{0},\ \ \text{then}\ \ \vec{a}\times\vec{b}=\vec{0}.$ Is the converse true? Justify your answer with an example.

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Answer

$\text{Given:}\ \ \text{Either}\ \vec{a}=\vec{0}\ \ \ \text{or}\ \ \ \vec{b}=\vec{0}$

$\therefore\ \ \Big|\vec{a}\Big|=\Big|\vec{0}\Big|=0\ \ \text{or}\ \ \ \Big|\vec{b}\Big|=\Big|\vec{0}\Big|=0\ \ \ ........\text{(i)}$

$\therefore\ \Big|\vec{a}\times\vec{b}\Big|=\Big|\vec{a}\Big|\Big|\vec{b}\Big|\sin\theta=0\ $ $\Rightarrow\ \ 0.\sin\theta=0\ \ \big[\text{Using eq.(i)}\big]$

$\therefore\ \vec{a}\times\vec{b}=\vec{0}\ \ \ \ \text{[By definition of zero vector]}$

But the converse is not true.

$\text{Let}\ \ \ \vec{a}=\hat{i}+\hat{j}+\hat{k}\ $ $\Rightarrow\ \ \big|\vec{a}\big|=\sqrt{1+1+1}=\sqrt{3}\neq0$

$\therefore\ \ \ \ \ \ \vec{a}\ \text{is a non-zero vector.}$

$\text{Let}\ \ \ \vec{b}=2\Big(\hat{i}+\hat{j}+\hat{k}\Big)=2\hat{i}+2\hat{j}+2\hat{k}\ $ $\Rightarrow\ \ \Big|\vec{b}\Big|=\sqrt{4+4+4}=\sqrt{12}=2\sqrt{3}\neq0$

$\therefore\ \ \ \ \ \vec{b}\ \text{is a non-zero vector.}$

$\text{But}\ \ \ \ \ \ \vec{a}\times\ \vec{b}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&1&1\\2&2&2\end{vmatrix}$

$\text{Taking 2 common from R}_{3}=\vec{a}\times\ \vec{b}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&1&1\\1&1&1\end{vmatrix}=\vec{0}\ \ \ $ $\big[\because\text{R}_2\ \text{and R}_{3}\ \text{are identical}\big]$

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