MHT CET · Physics · Mathematics in Physics
The unit vector \((a \hat{\imath}+b \hat{j})\) is perpendicular to \((\hat{\imath}+\hat{\jmath})\). The value of \(^{\prime} b^{\prime}\) is
- A \(+\frac{1}{\sqrt{3}}\)
- B \(-\frac{1}{\sqrt{3}}\)
- C \(+\frac{1}{2}\)
- D \(-\frac{1}{\sqrt{2}}\)
Answer & Solution
Correct Answer
(D) \(-\frac{1}{\sqrt{2}}\)
Step-by-step Solution
Detailed explanation
\( \begin{array}{l} \hat{\mathrm{n}}=a \hat{\mathrm{i}}+b \hat{\mathrm{j}} \\ \overrightarrow{\mathrm{r}}=\hat{\mathrm{i}}+\hat{\mathrm{j}} \end{array} \)
If \(\hat{n}\) is perpendicular to the vector \(\vec{r}=\hat{i}+\hat{j}\)
\(\hat{n} \cdot \vec{r}=0\)
\(a+b=0\)
\(b=-a . \ldots . .(1)\)
The magnitude of unity vector is \(1 .\)
\(\sqrt{a^{2}+b^{2}}=1\) \(\sqrt{a^{2}+a^{2}}=\sqrt{2} a=1(\) from equation 1) \(a=\frac{1}{\sqrt{2}}\)
\(a=\frac{1}{\sqrt{2}}\)
and from equation (1), we get
\(b=-\frac{1}{\sqrt{2}}\)
The correct option is D.
If \(\hat{n}\) is perpendicular to the vector \(\vec{r}=\hat{i}+\hat{j}\)
\(\hat{n} \cdot \vec{r}=0\)
\(a+b=0\)
\(b=-a . \ldots . .(1)\)
The magnitude of unity vector is \(1 .\)
\(\sqrt{a^{2}+b^{2}}=1\) \(\sqrt{a^{2}+a^{2}}=\sqrt{2} a=1(\) from equation 1) \(a=\frac{1}{\sqrt{2}}\)
\(a=\frac{1}{\sqrt{2}}\)
and from equation (1), we get
\(b=-\frac{1}{\sqrt{2}}\)
The correct option is D.
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