65. Let a→=i^+4j^+2k^, b→=3i^−2j^+7k^ and c→=2i^−j^+4k^. Find a vector d→ which is perpendicular to both a→ and b→ and c→.d→ = 15

Given, a→=i^+4j^+2k^b→=3i^−2j^+7k^c→=2i^−j^+4k^ Let, d→=d1i^+d2j^+d3k^ Since, d→ is perpendicular to both a→&b→ d→.a→=0⇒d1+d24+d32=0−−−−−(1)d→.b→=0⇒d13+d2(−2)+d3(7)=0⇒d13−2d2+7d3=0−−−−−(2) We know, c→.d→=15⇒2d1−d2+4d3=15−−−−−(3)From,(1)d1+4d2+2d3=0d1=−4d2−2d3 Putting this value in (3) we get ⇒2(−4d2−2d3)−d2+4d3=15⇒−8d2−4d3−d2+4d3=15⇒−9d2=15⇒d2=159=−53 Putting d1&d2 value in (2), we get ⇒3d1−2d2+7d3=0⇒3(−4d2−2d3)−2(−53)+7d3=0⇒−12*(−53)−6d3+103+7d3=0⇒20+d3+103=0⇒d3=−20−103=−60−103=−703Now,d1=−4*−53−2*−703=203+1403=1603∴d1=1603,d2=−53,d3=−703d→=1603i^−53j^−703k^=13(160i^−5j^−70k^) ∴ The reQ.uired vector is 13(160i^−5j^−70k^)

65. Let $\vec{a} = \hat{i} + 4 \hat{j} + 2 \hat{k}, \vec{b} = 3 \hat{i} - 2 \hat{j} + 7 \hat{k} a n d \vec{c} = 2 \hat{i} - \hat{j} + 4 \hat{k} .$ Find a vector $\vec{d}$ which is perpendicular to both $\vec{a}$ and $\vec{b}$ and $\vec{c} . \vec{d}$ = 15

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Vishal Baghel | Contributor-Level 10

8 months ago

Given,

$\begin{array}{l} \vec{a} = \hat{i} + 4 \hat{j} + 2 \hat{k} \\ \vec{b} = 3 \hat{i} - 2 \hat{j} + 7 \hat{k} \\ \vec{c} = 2 \hat{i} - \hat{j} + 4 \hat{k} \end{array}$

Let, $\vec{d} = d_{1} \hat{i} + d_{2} \hat{j} + d_{3} \hat{k}$

Since, $\vec{d}$ is perpendicular to both $\vec{a} & \vec{b}$

$\begin{array}{l} \vec{d} . \vec{a} = 0 \\ \Rightarrow d_{1} + d_{2} 4 + d_{3} 2 = 0 - - - - - (1) \\ \vec{d} . \vec{b} = 0 \\ \Rightarrow d_{1} 3 + d_{2} (- 2) + d_{3} (7) = 0 \\ \Rightarrow d_{1} 3 - 2 d_{2} + 7 d_{3} = 0 - - - - - (2) \end{array}$

We know,

$\begin{array}{l} \vec{c} . \vec{d} = 15 \\ \Rightarrow 2 d_{1} - d_{2} + 4 d_{3} = 15 - - - - - (3) \\ F r o m, (1) \\ d_{1} + 4 d_{2} + 2 d_{3} = 0 \\ d_{1} = - 4 d_{2} - 2 d_{3} \end{array}$

Putting this value in (3) we get

$\begin{array}{l} \Rightarrow 2 (- 4 d_{2} - 2 d_{3}) - d_{2} + 4 d_{3} = 15 \\ \Rightarrow - 8 d_{2} - 4 d_{3} - d_{2} + 4 d_{3} = 15 \\ \Rightarrow - 9 d_{2} = 15 \\ \Rightarrow d_{2} = \frac{15}{9} = - \frac{5}{3} \end{array}$

Putting $d_{1} & d_{2}$ value in (2), we get

$\begin{array}{l} \Rightarrow 3 d_{1} - 2 d_{2} + 7 d_{3} = 0 \\ \Rightarrow 3 (- 4 d_{2} - 2 d_{3}) - 2 (- \frac{5}{3}) + 7 d_{3} = 0 \\ \Rightarrow - 12 * (- \frac{5}{3}) - 6 d_{3} + \frac{10}{3} + 7 d_{3} = 0 \\ \Rightarrow 20 + d_{3} + \frac{10}{3} = 0 \\ \Rightarrow d_{3} = - 20 - \frac{10}{3} = \frac{- 60 - 10}{3} = \frac{- 70}{3} \\ N o w, d_{1} = - 4 * - \frac{5}{3} - 2 * - \frac{70}{3} \\ = \frac{20}{3} + \frac{140}{3} = \frac{160}{3} \\ ∴ d_{1} = \frac{160}{3}, d_{2} = - \frac{5}{3}, d_{3} = \frac{- 70}{3} \\ \vec{d} = \frac{160}{3} \hat{i} - \frac{5}{3} \hat{j} \frac{- 70}{3} \hat{k} = \frac{1}{3} (160 \hat{i} - 5 \hat{j} - 70 \hat{k}) \end{array}$