Question

If the projection of the vector \(\hat{i} + 2\hat{j} + \hat{k}\) on the sum of the two vectors \(2\hat{i} + 4\hat{j} - 5\hat{k}\) and \(-\lambda\hat{i} + 2\hat{j} + 3\hat{k}\) is 1, then \(\lambda\) is equal to ___________.

Hint:

Remember the formula for the projection of \(\vec{a}\) on \(\vec{b}\) is \(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}\). A common mistake is to divide by \(|\vec{a}|\) or forget the magnitude in the denominator. Squaring both sides of an equation can introduce extraneous solutions, so it's a good practice to check if the solution satisfies the equation before squaring.

Answer 1

Correct Answer: 5

Step 1: Understanding the Question
We are asked to find the value of \(\lambda\) given that the projection of one vector onto another is 1.
Step 2: Key Formula or Approach
The projection of a vector \(\vec{a}\) onto a vector \(\vec{b}\) is given by the formula: \[ \text{proj}_{\vec{b}}\vec{a} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} \] Step 3: Detailed Explanation
Let \(\vec{a} = \hat{i} + 2\hat{j} + \hat{k}\). Let \(\vec{v_1} = 2\hat{i} + 4\hat{j} - 5\hat{k}\) and \(\vec{v_2} = -\lambda\hat{i} + 2\hat{j} + 3\hat{k}\). Let \(\vec{b}\) be the sum of \(\vec{v_1}\) and \(\vec{v_2}\). \[ \vec{b} = \vec{v_1} + \vec{v_2} = (2\hat{i} + 4\hat{j} - 5\hat{k}) + (-\lambda\hat{i} + 2\hat{j} + 3\hat{k}) \] \[ \vec{b} = (2-\lambda)\hat{i} + (4+2)\hat{j} + (-5+3)\hat{k} = (2-\lambda)\hat{i} + 6\hat{j} - 2\hat{k} \] We are given that the projection of \(\vec{a}\) on \(\vec{b}\) is 1. \[ \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = 1 \implies \vec{a} \cdot \vec{b} = |\vec{b}| \] First, calculate the dot product \(\vec{a} \cdot \vec{b}\): \[ \vec{a} \cdot \vec{b} = (\hat{i} + 2\hat{j} + \hat{k}) \cdot ((2-\lambda)\hat{i} + 6\hat{j} - 2\hat{k}) \] \[ = 1(2-\lambda) + 2(6) + 1(-2) = 2 - \lambda + 12 - 2 = 12 - \lambda \] Next, calculate the magnitude of \(\vec{b}\): \[ |\vec{b}| = \sqrt{(2-\lambda)^2 + 6^2 + (-2)^2} = \sqrt{(2-\lambda)^2 + 36 + 4} = \sqrt{(2-\lambda)^2 + 40} \] Now, set \(\vec{a} \cdot \vec{b} = |\vec{b}|\): \[ 12 - \lambda = \sqrt{(2-\lambda)^2 + 40} \] Square both sides (we must have \(12-\lambda>0\)): \[ (12 - \lambda)^2 = (2-\lambda)^2 + 40 \] \[ 144 - 24\lambda + \lambda^2 = 4 - 4\lambda + \lambda^2 + 40 \] \[ 144 - 24\lambda = 44 - 4\lambda \] \[ 100 = 20\lambda \] \[ \lambda = 5 \] We must check our assumption \(12-\lambda>0\). For \(\lambda=5\), \(12-5=7>0\), so the solution is valid.
Step 4: Final Answer
The value of \(\lambda\) is 5.