The vector equation of the plane passing through the point i^+j^−2k^,2i^−j^+k^ and i^+2j^+k^, is

Let A, B, C be the points with position vectors i^+j^−2k^,2i^−j^+k^ and i^+2j^+k^ respectively. Then,AB→=i^−2j^+3k^ and AB→=−i^+3j^+0k^A vector normal to the plane containing points A, B, and C, isn→=AB→×AC→=i^ j^ k^1 −2 3−1 3 0=−9i^−3j^+k^The required plane passes through the point having position vector a→=i^+j^−2k^ and and is normal to the vector n→=−9i^−3j^+k^ So, its vector equation is⇒ r→⋅n→=a→⋅n→⇒i→⋅(9i^+3j^−k^)=14

The vector equation of the plane passing through the point i^+j^−2k^,2i^−j^+k^ and i^+2j^+k^, is

Let A, B, C be the points with position vectors i^+j^−2k^,2i^−j^+k^ and i^+2j^+k^ respectively. Then,AB→=i^−2j^+3k^ and AB→=−i^+3j^+0k^A vector normal to the plane containing points A, B, and C, isn→=AB→×AC→=i^ j^ k^1 −2 3−1 3 0=−9i^−3j^+k^The required plane passes through the point having position vector a→=i^+j^−2k^ and and is normal to the vector n→=−9i^−3j^+k^ So, its vector equation is⇒ r→⋅n→=a→⋅n→⇒i→⋅(9i^+3j^−k^)=14

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

The vector equation of the plane passing through the point $\hat{i} + \hat{j} - 2 \hat{k}, 2 \hat{i} - \hat{j} + \hat{k}$ and $\hat{i} + 2 \hat{j} + \hat{k}, is$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

None of these

$\vec{r} \cdot (9 \hat{i} + 3 \hat{j} - \hat{k}) = 14$

$\vec{r} \cdot (3 \hat{i} + 9 \hat{j} - \hat{k}) = 14$

$\vec{r} \cdot (9 \hat{i} + 3 \hat{j} - \hat{k}) = - 14$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Let A, B, C be the points with position vectors $\hat{i} + \hat{j} - 2 \hat{k}, 2 \hat{i} - \hat{j} + \hat{k}$ and $\hat{i} + 2 \hat{j} + \hat{k}$ respectively. Then,

$\vec{A B} = \hat{i} - 2 \hat{j} + 3 \hat{k}$ and $\vec{A B} = - \hat{i} + 3 \hat{j} + 0 \hat{k}$

A vector normal to the plane containing points A, B, and C, is

$\vec{n} = \vec{A B} \times \vec{A C} = |\begin{array}{rrr} \hat{i} \hat{j} \hat{k} \\ 1 - 2 3 \\ - 1 3 0 \end{array}| = - 9 \hat{i} - 3 \hat{j} + \hat{k}$

The required plane passes through the point having position vector $\vec{a} = \hat{i} + \hat{j} - 2 \hat{k}$ and and is normal to the vector $\vec{n} = - 9 \hat{i} - 3 \hat{j} + \hat{k}$