Question
Download Solution PDFDetermine \(\left| {\overrightarrow a } \right| \rm and\left| {\overrightarrow b } \right|,if(\overrightarrow a + \overrightarrow b ).(\overrightarrow a - \overrightarrow b ) = 8\ and\left| {\overrightarrow a } \right| = 8\left| {\overrightarrow b } \right|\)?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFGiven:
\((\overrightarrow a + \overrightarrow b ).(\overrightarrow a - \overrightarrow b ) = 8\ \)
\(\left| {\overrightarrow a } \right| = 8\left| {\overrightarrow b } \right|\)
Calculation:
We have,
\((\overrightarrow a + \overrightarrow b ).(\overrightarrow a - \overrightarrow b ) = 8\ \)
⇒ \(\vec a.\vec a \ - \vec a . \vec b + \vec b . \vec a - \vec b. \vec b = 8 \)
⇒ \(\left|\vec a \right|^2 - \vec a .\vec b + \vec a . \vec b - \left|\vec b \right|^2 = 8\) [∵ \(\vec b. \vec a = \vec a . \vec b\)]
⇒ \(\left| \vec a \right|^2 - \left| \vec b \right|^2 = 8\)
⇒ \((8\left| \vec b \right|)^2 - \left| \vec b \right|^2 = 8\) [\(\left| {\overrightarrow a } \right| = 8\left| {\overrightarrow b } \right|\)]
⇒ \(64 \left| \vec b \right|^2 - \left| \vec b \right|^2 = 8\)
⇒ \(63 \left| \vec b \right|^2 = 8\)
⇒ \(\left| \vec b \right|^2 = \frac{8}{63}\)
⇒ \(\left| \vec b \right| = \sqrt {\frac{8}{63}} \)
⇒ \(\left| \vec b \right| = { \frac{2\sqrt2}{3\sqrt7}} \)
⇒ \(\left| {\overrightarrow a } \right| = 8\left| {\overrightarrow b } \right|\)
Put the value of \(\vec b \)
⇒ \(\left| \vec a \right| = 8 \times \frac{2\sqrt2}{3\sqrt7}\)
⇒ \(\left| \vec a \right| = \frac{16\sqrt2}{3\sqrt7}\)
∴ \(\left| \vec a \right| = \frac{16\sqrt2}{3\sqrt7}\) and \(\left| \vec b \right| = { \frac{2\sqrt2}{3\sqrt7}} \)
Last updated on May 26, 2025
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