What is the projection of $(3 i + 2 j - 6 k)$ onto $(3 i - 4 j + 4 k)$

14

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- 14

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12

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none ofthese.

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Solution

The correct option is D none ofthese.
$\begin{matrix} G i v e n \to a = (3 i + 2 j - 6 k) a n d \to b (3 i - 4 j + 4 k) N o w, o j_{\to b}^{\to a} (\frac{\to a . \to b}{∣ ∣ \to b ∣ ∣}) \frac{\to b}{∣ ∣ \to b ∣ ∣} B y t a k i n g t h e d o t p r o d u c t o f t h e t w o v e c t o r s, \to a =< 3, 2, - 6 > a n d \to b =< 3, - 4, 4 > \to a . \to b =< 3, 2, - 4 > . < 3, - 4, 4 > \Rightarrow (3.3) + (2. - 4) + (- 6.4) \Rightarrow 9 - 8 - 24 = - 23 | \to b | = \sqrt{{(b_{x})}^{2} + {(b_{y})}^{2} + {(b_{z})}^{2}} = \sqrt{3^{2} + {(- 4)}^{2} + {(4)}^{2}} = \sqrt{9 + 16 + 16} = \sqrt{41} N o w, p r o j_{\to b}^{\to a} = \frac{- 23}{\sqrt{41}} . \frac{< 3, - 4, 4 >}{\sqrt{41}} \Rightarrow \frac{- 23 < 3, - 4, 4 >}{41} \Rightarrow \frac{- 23}{41} < 3, - 4, 4 > \Rightarrow< \frac{69}{41}, \frac{92}{41}, \frac{- 92}{41} > w h i c h i s t h e p r o j e c t i o n o f t h e l i n e . \end{matrix}$
hence, option $D$ is the required answer.

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Similar questions

A unit vector perpendicular to the plane determined by the vectors $4^i + 3^j -^k a n d 2^i - 6^j - 3^k$

\to r = 3 i + 2 j - 4 k + λ (i + 2 j + 2 k)

\to r = (5 j - 2 k) + μ (3 i + 2 j + 6 k)

\to r = 3 i + 2 j - 4 k + λ (i + 2 j + 2 k)

\to r = 5 i - 2 k + μ (3 i + 2 j + 6 k)

\vec{r} = (4 \hat{i} - \hat{j}) + λ (\hat{i} + 2 \hat{j} - 2 \hat{k}) and \vec{r} = \hat{i} - \hat{j} + 2 \hat{k} - μ (2 \hat{i} + 4 \hat{j} - 4 \hat{k})

\vec{r} = (3 \hat{i} + 2 \hat{j} - 4 \hat{k}) + λ (\hat{i} + 2 \hat{j} + 2 \hat{k}) and \vec{r} = (5 \hat{j} - 2 \hat{k}) + μ (3 \hat{i} + 2 \hat{j} + 6 \hat{k})

\vec{r} = λ (\hat{i} + \hat{j} + 2 \hat{k}) and \vec{r} = 2 \hat{j} + μ \{(\sqrt{3} - 1) \hat{i} - (\sqrt{3} + 1) \hat{j} + 4 \hat{k}\}

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