Let vec a = 2hat i + lambda_1hat j + 3hat k , vec b = 4hat i + (3-lambda_2)hat j + 6hat k and vec c = 3hat i + 6hat j + (lambda_3-1)hat k be three vectors such that vec b = 2vec a and vec a is perpendicular to vec c . Then a possible value of (lambda_1, lambda_2, lambda_3) is:

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EEJ MAIN Mathematics QUESTION #7479

Question 1

Let \(\vec{a} = 2\hat{i} + \lambda_1\hat{j} + 3\hat{k}\), \(\vec{b} = 4\hat{i} + (3-\lambda_2)\hat{j} + 6\hat{k}\) and \(\vec{c} = 3\hat{i} + 6\hat{j} + (\lambda_3-1)\hat{k}\) be three vectors such that \(\vec{b} = 2\vec{a}\) and \(\vec{a}\) is perpendicular to \(\vec{c}\). Then a possible value of \((\lambda_1, \lambda_2, \lambda_3)\) is:

(1, 3, 1)
\((-\frac{1}{2}, 4, 0)\)✔️
\((\frac{1}{2}, 4, -2)\)
(1, 5, 1)

Correct Answer Explanation

From \(\vec{b} = 2\vec{a}\): \(4\hat{i}+(3-\lambda_2)\hat{j}+6\hat{k} = 4\hat{i}+2\lambda_1\hat{j}+6\hat{k}\). Comparing: \(3-\lambda_2 = 2\lambda_1\), so \(\lambda_2 = 3-2\lambda_1\). From \(\vec{a} \perp \vec{c}\): \(2(3) + \lambda_1(6) + 3(\lambda_3-1) = 0\), giving \(6+6\lambda_1+3\lambda_3-3 = 0\), so \(6\lambda_1+3\lambda_3 = -3\), thus \(2\lambda_1+\lambda_3 = -1\). Testing option B: \(\lambda_1 = -\frac{1}{2}\), \(\lambda_2 = 4\), \(\lambda_3 = 0\). Check: \(\lambda_2 = 3-2(-\frac{1}{2}) = 3+1 = 4\) ✓. And \(2(-\frac{1}{2})+0 = -1\) ✓.

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