MHT CET · Physics · Mathematics in Physics
The angle made by a vector \(\overrightarrow{\mathrm{B}}=3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}\) with \(\mathrm{y}\) -axis is
- A \(\cos ^{-1}\left(\frac{5}{\sqrt{23}}\right)\)
- B \(\cos ^{-1}\left(\frac{4}{\sqrt{11}}\right)\)
- C \(\cos ^{-1}\left(\frac{3}{\sqrt{17}}\right)\)
- D \(\cos ^{-1}\left(\frac{2}{\sqrt{29}}\right)\)
Answer & Solution
Correct Answer
(D) \(\cos ^{-1}\left(\frac{2}{\sqrt{29}}\right)\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{l}
\vec{B}=3 \hat{\imath}+2 \hat{\jmath}+4 \hat{k} \\
y \text {-axis }=\hat{\jmath}
\end{array}\)
\(\angle\) made with y axis \(\cos ^{-1}(\sqrt{2}\)
\(=\cos ^{-1}\left(\frac{2}{\sqrt{29}}\right)\)
\vec{B}=3 \hat{\imath}+2 \hat{\jmath}+4 \hat{k} \\
y \text {-axis }=\hat{\jmath}
\end{array}\)
\(\angle\) made with y axis \(\cos ^{-1}(\sqrt{2}\)
\(=\cos ^{-1}\left(\frac{2}{\sqrt{29}}\right)\)
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