Question
Dot product of a vector with $\hat{\text{i}}+\hat{\text{j}}-3\hat{\text{k}},\hat{\text{i}}+3\hat{\text{j}}-2\hat{\text{k}}$ and $2\hat{\text{i}}+\hat{\text{j}}+4\hat{\text{k}}$ are 0, 5 and 8 respectively. Find the vector.
✓
Answer
Let $\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}}$ be the required vector.
Given that
$\big(\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}}\big).\big(\hat{\text{i}}+\hat{\text{j}}-3\hat{\text{k}}\big)=0$
$\Rightarrow\text{a+b}-3\text{c}=0\dots(1)$
$\big(\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}}\big).\big(\hat{\text{i}}+3\hat{\text{j}}-2\hat{\text{k}}\big)=5$
$\Rightarrow\text{a}+3\text{b}-2\text{c}=5\dots(2)$
$\big(\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}}\big).\big(2\hat{\text{i}}+\hat{\text{j}}+4\hat{\text{k}}\big)=5$
$\Rightarrow2\text{a+b}+4\text{c}=8\dots(3)$
Solving (1), (2) and (3), we get
$\text{a}=1,\text{b}=2,\text{c}=1$
So, $\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}}=\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}$
Given that
$\big(\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}}\big).\big(\hat{\text{i}}+\hat{\text{j}}-3\hat{\text{k}}\big)=0$
$\Rightarrow\text{a+b}-3\text{c}=0\dots(1)$
$\big(\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}}\big).\big(\hat{\text{i}}+3\hat{\text{j}}-2\hat{\text{k}}\big)=5$
$\Rightarrow\text{a}+3\text{b}-2\text{c}=5\dots(2)$
$\big(\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}}\big).\big(2\hat{\text{i}}+\hat{\text{j}}+4\hat{\text{k}}\big)=5$
$\Rightarrow2\text{a+b}+4\text{c}=8\dots(3)$
Solving (1), (2) and (3), we get
$\text{a}=1,\text{b}=2,\text{c}=1$
So, $\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}}=\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}$
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