Question 62 of 68advanced🔧 ApplyNumerical4 marks

The scalar product of the vector î + ĵ + k̂ with a unit vector along the sum of vectors 2î + 4ĵ - 5k̂ and λî + 2ĵ + 3k̂ is equal to one. Find the value of λ.

Correct Answer

λ = 1

Exercise: MISCELLANEOUS EXERCISE ON CHAPTER 10 | Q: 13 | (Page 35)
For More Understanding

Explanation

This question tests the concepts of vector addition, finding unit vectors, and scalar product. The key steps involve: (1) adding two vectors to get the sum vector, (2) finding the unit vector along this sum, (3) computing the scalar product with the given vector, and (4) equating it to 1 to solve for λ.

Solution Steps

  1. Step 1: Let the two vectors be a⃗=2i^+4j^−5k^\vec{a} = 2\hat{i} + 4\hat{j} - 5\hat{k} and b⃗=λi^+2j^+3k^\vec{b} = \lambda\hat{i} + 2\hat{j} + 3\hat{k}. Their sum is a⃗+b⃗=(2+λ)i^+6j^−2k^\vec{a} + \vec{b} = (2+\lambda)\hat{i} + 6\hat{j} - 2\hat{k}.

  2. Step 2: The magnitude of the sum vector is ∣a⃗+b⃗∣=(2+λ)2+62+(−2)2=(2+λ)2+40|\vec{a} + \vec{b}| = \sqrt{(2+\lambda)^2 + 6^2 + (-2)^2} = \sqrt{(2+\lambda)^2 + 40}.

  3. Step 3: The unit vector along the sum is (2+λ)i^+6j^−2k^(2+λ)2+40\frac{(2+\lambda)\hat{i} + 6\hat{j} - 2\hat{k}}{\sqrt{(2+\lambda)^2 + 40}}.

  4. Step 4: The scalar product of i^+j^+k^\hat{i} + \hat{j} + \hat{k} with this unit vector equals (2+λ)+6+(−2)(2+λ)2+40=λ+6(2+λ)2+40\frac{(2+\lambda) + 6 + (-2)}{\sqrt{(2+\lambda)^2 + 40}} = \frac{\lambda + 6}{\sqrt{(2+\lambda)^2 + 40}}.

  5. Step 5: Given that this scalar product equals 1: λ+6(2+λ)2+40=1\frac{\lambda + 6}{\sqrt{(2+\lambda)^2 + 40}} = 1. Squaring both sides: (λ+6)2=(2+λ)2+40(\lambda + 6)^2 = (2+\lambda)^2 + 40. Simplifying: λ2+12λ+36=λ2+4λ+44\lambda^2 + 12\lambda + 36 = \lambda^2 + 4\lambda + 44, which gives 8λ=88\lambda = 8, so λ = 1.

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