Show that the points whose position vectors 4i- - 5j- - k-3i- - 9j- - 4k- and - 4i- - 4j- - 4k- are coplanar

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Standard XII

Mathematics

Collinear Vectors

Show that the...

Question

Show that the points whose position vectors $4 \to i + 5 \to j + \to k, 3 \to i + 9 \to j + 4 \to k$ and $- 4 \to i + 4 \to j + 4 \to k$ are coplanar

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Solution

$\begin{matrix} Solution: calculate a scalar triple product of vectors ¯ ¯ ¯ a \cdot [¯ ¯ b \times ¯ ¯ c] = ∣ ∣ ∣ ∣ \begin{matrix} 4 & 5 & 1 3 & 9 & 4 - 4 & 4 & 4 \end{matrix} ∣ ∣ ∣ ∣ = 0 Answer: vectors are coplanar as their scalar triple product is zero. \end{matrix}$

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Similar questions

Q. P,Q,R and S are four points with the position vectors

3 \to i - 4 \to j + 5 \to k, 4 \to k, - 4 \to i + 5 \to j + \to k a n d - 3 \to i + 4 \to j + 3 \to k

respectively. Then the line PQ meets the line RS at the point

P,Q,R and S are four points with the position vectors, $3 \to i - 4 \to j + 5 \to k, 4 \to k, - 4 \to i + 5 \to j + \to k and - 3 \to i + 4 \to j + 3 \to k$ respectively. Then the line PQ meets the line RS at the point

Q. Show that the points whose position vectors are

4 \to i - 3 \to j + \to k, 2 \to i - 4 \to j + 5 \to k, \to i - \to j

from a right angled triangle.

Q. The vectors

2 \to i + 3 \to j + 4 \to k, 3 \to i + 2 \to j - 3 \to k

, and

5 \to i + 5 \to j + \to k

form:

Q. Show that the vectors

2 \to i - \to j + \to k

\to i - 3 \to j - 5 \to k

and

- 3 \to i + 4 \to j + 4 \to k

are the sides of a right angled triangle.

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Show that the points whose position vectors 4→i+5→j+→k,3→i+9→j+4→k and −4→i+4→j+4→k are coplanar

Solution: calculate a scalar triple product of vectors ¯¯¯a⋅[¯¯b×¯¯c]=∣∣ ∣∣451394−444∣∣ ∣∣=0 Answer: vectors are coplanar as their scalar triple product is zero.

Show that the points whose position vectors $4 \to i + 5 \to j + \to k, 3 \to i + 9 \to j + 4 \to k$ and $- 4 \to i + 4 \to j + 4 \to k$ are coplanar

$\begin{matrix} Solution: calculate a scalar triple product of vectors ¯ ¯ ¯ a \cdot [¯ ¯ b \times ¯ ¯ c] = ∣ ∣ ∣ ∣ \begin{matrix} 4 & 5 & 1 3 & 9 & 4 - 4 & 4 & 4 \end{matrix} ∣ ∣ ∣ ∣ = 0 Answer: vectors are coplanar as their scalar triple product is zero. \end{matrix}$