Question:
The scalar product of the vector \(\hat{i} + \hat{j} + \hat{k}\) with a unit vector along the sum of vectors \(2\hat{i} + 4\hat{j} – 5\hat{k}\) and \(\lambda\hat{i} + 2\hat{j} + 3\hat{k}\) is equal to one. Find the value of \(\lambda\).
सदिश \(\hat{i} + \hat{j} + \hat{k}\) का सदिशों \(2\hat{i} + 4\hat{j} – 5\hat{k}\) और \(\lambda\hat{i} + 2\hat{j} + 3\hat{k}\) के योग के अनुदिश मात्रक सदिश के साथ अदिश गुणनफल 1 के बराबर है। \(\lambda\) का मान ज्ञात कीजिए।
Options:
- \(\lambda = -2\)
- \(\lambda = 1\)
- \(\lambda = 2\)
- \(\lambda = -1\)
- \(\lambda = -2\)
- \(\lambda = 1\)
- \(\lambda = 2\)
- \(\lambda = -1\)
Correct Answer: b)
Explanation:
Sum vector \(\vec{S} = (2+\lambda)\hat{i} + 6\hat{j} – 2\hat{k}\).
\((\hat{i}+\hat{j}+\hat{k}) \cdot \dfrac{\vec{S}}{|\vec{S}|} = 1 \implies \lambda + 6 = \sqrt{(\lambda+2)^2 + 36 + 4}\).
Squaring: \(\lambda^2 + 12\lambda + 36 = \lambda^2 + 4\lambda + 44 \implies 8\lambda = 8 \implies \lambda = 1\).
योग सदिश \(\vec{S} = (2+\lambda)\hat{i} + 6\hat{j} – 2\hat{k}\)।
प्रश्नानुसार: \(\dfrac{\lambda + 6}{\sqrt{(\lambda+2)^2 + 40}} = 1 \implies 8\lambda = 8 \implies \lambda = 1\)।