\( a\) and \( b\) are non-zero non-collinear vectors such that \(| a| = 2, a. b = 1\) and the angle between \( a\) and \( b\) is \( π 3\) . If <span...

\( a · b = | a || b |cosπ/3 = 2| b | · 1/2 = | b | = 1\) \(| a × b | 2 = | a | 2| b | 2 - ( a · b ) 2 = 4 - 1 = 3\) Since \( r + 2 a - 10 b = λ( a × b )\), dotting with \( a × b \): \(λ| a × b | 2 = 6\). \(3λ = 6 λ = 2\) Answer: (B) 2

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Question 22

Mathematics Scalar and Vector Products Hard

\(\vec a\) and \(\vec b\) are non-zero non-collinear vectors such that \(|\vec a| = 2, \vec a.\vec b = 1\) and the angle between \(\vec a\) and  \(\vec b\) is \(\frac \pi 3\). If \(\vec r\) is any vector satisfying \(\vec r.\vec a = 2,\;\vec r.\vec b = 8,\;(\vec r + 2\vec a - 10\vec b).(\vec a \times \vec b) = 6\) and \(\vec r + 2\vec a - 10 \vec b = λ(\vec a \times \vec b),\) then λ =

(A) \(\frac 1 2\)

(B) 2

(C) \(\frac 4 {\sqrt 3}\)

(D) 3

✓ Correct Answer: Option B Verified Solution

📘 Step-by-Step Solution & Detailed Explanation:

\(\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\frac{\pi}{3} = 2|\vec{b}| \cdot \frac{1}{2} = |\vec{b}| = 1\)

\(|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2|\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2 = 4 - 1 = 3\)

Since \(\vec{r} + 2\vec{a} - 10\vec{b} = \lambda(\vec{a} \times \vec{b})\), dotting with \(\vec{a} \times \vec{b}\): \(\lambda|\vec{a} \times \vec{b}|^2 = 6\).

\(3\lambda = 6 \Rightarrow \lambda = 2\)

Answer: (B) 2

Question 22

📘 Step-by-Step Solution & Detailed Explanation:

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