Question ID: #883
Let $\vec{a}=2\hat{i}-\hat{j}-\hat{k}$, $\vec{b}=\hat{i}+3\hat{j}-\hat{k}$ and $\vec{c}=2\hat{i}+\hat{j}+3\hat{k}$. Let $\vec{v}$ be the vector in the plane of the vectors $\vec{a}$ and $\vec{b}$, such that the length of its projection on the vector $\vec{c}$ is equal to $\frac{1}{\sqrt{14}}$. Then $|\vec{v}|$ is equal to
- (1) $\frac{\sqrt{21}}{2}$
- (2) 13
- (3) $\frac{\sqrt{35}}{2}$
- (4) 7
Solution:
Since $\vec{v}$ is in the plane of $\vec{a}$ and $\vec{b}$:
$$\vec{v} = x\vec{a} + y\vec{b} = x(2\hat{i}-\hat{j}-\hat{k}) + y(\hat{i}+3\hat{j}-\hat{k})$$
$$\vec{v} = (2x+y)\hat{i} + (3y-x)\hat{j} – (x+y)\hat{k}$$
Projection of $\vec{v}$ on $\vec{c}$ is $\frac{|\vec{v} \cdot \vec{c}|}{|\vec{c}|} = \frac{1}{\sqrt{14}}$
$$|\vec{c}| = \sqrt{2^2 + 1^2 + 3^2} = \sqrt{4+1+9} = \sqrt{14}$$
$$\vec{v} \cdot \vec{c} = 2(2x+y) + 1(3y-x) + 3(-x-y) = 4x + 2y + 3y – x – 3x – 3y = 2y$$
$$\frac{|2y|}{\sqrt{14}} = \frac{1}{\sqrt{14}} \Rightarrow |2y| = 1 \Rightarrow y = \pm \frac{1}{2}$$
$$|\vec{v}|^2 = (2x+y)^2 + (3y-x)^2 + (x+y)^2$$
$$|\vec{v}|^2 = (4x^2+y^2+4xy) + (9y^2+x^2-6xy) + (x^2+y^2+2xy)$$
$$|\vec{v}|^2 = 6x^2 + 11y^2 = 6x^2 + 11\left(\frac{1}{4}\right) = 6x^2 + \frac{11}{4}$$
$$|\vec{v}| = \frac{\sqrt{24x^2+11}}{2}$$
To find $x$ one more condition required.
Ans. (Bonus)
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