Question
Show that the four points having position vectors $6\hat{\text{i}}-7\hat{\text{j}},\ 16\hat{\text{i}}-19\hat{\text{j}}-4\hat{\text{k}},\ 3\hat{\text{j}}-6\hat{\text{k}},\ 2\hat{\text{i}}-5\hat{\text{j}}+10\hat{\text{k}}$ are coplanar.
✓
Answer
Let the given four points be P, Q, R and S respectively. Three points are coplanar if the vectors $\overrightarrow{\text{PQ}},\ \overrightarrow{\text{PR}}\text{ and }\overrightarrow{\text{PS}}$ are coplanar. These vectors are coplanar if one of them can be expressed as a linear combination of the other two. So, let$\overrightarrow{\text{PQ}}=\text{x}\overrightarrow{\text{PR}}+\text{y}\overrightarrow{\text{PS}}$
$10\hat{\text{i}}-12\hat{\text{j}}-4\hat{\text{k}}\\=\text{x}\big(-6\hat{\text{i}}+10\hat{\text{j}}-6\hat{\text{k}}\big)+\text{y}\big(-4\hat{\text{i}}+2\hat{\text{j}}+10\hat{\text{k}}\big)$
$\Rightarrow10\hat{\text{i}}-12\hat{\text{j}}-4\hat{\text{k}}\\=\hat{\text{i}}\big(-6{\text{x}}-4\text{y}\big)+\hat{\text{j}}\big(10{\text{x}}+2{\text{y}}\big)+\hat{\text{k}}\big(-6\text{x}+10\text{y}\big)$
$\Rightarrow-6\text{x}-4\text{y}=10,\ 10\text{x}+2\text{y}=-12$ and $-6\text{x}+10\text{y}=-4$ [Equating coefficients of $\hat{\text{i}},\ \hat{\text{j}},\ \hat{\text{k}}$ on both sides]
Solving the first of these three equations, we get x = -1 and y = -1. These values also satisfy the thied equation. Hence, the given four points are coplanar.
$10\hat{\text{i}}-12\hat{\text{j}}-4\hat{\text{k}}\\=\text{x}\big(-6\hat{\text{i}}+10\hat{\text{j}}-6\hat{\text{k}}\big)+\text{y}\big(-4\hat{\text{i}}+2\hat{\text{j}}+10\hat{\text{k}}\big)$
$\Rightarrow10\hat{\text{i}}-12\hat{\text{j}}-4\hat{\text{k}}\\=\hat{\text{i}}\big(-6{\text{x}}-4\text{y}\big)+\hat{\text{j}}\big(10{\text{x}}+2{\text{y}}\big)+\hat{\text{k}}\big(-6\text{x}+10\text{y}\big)$
$\Rightarrow-6\text{x}-4\text{y}=10,\ 10\text{x}+2\text{y}=-12$ and $-6\text{x}+10\text{y}=-4$ [Equating coefficients of $\hat{\text{i}},\ \hat{\text{j}},\ \hat{\text{k}}$ on both sides]
Solving the first of these three equations, we get x = -1 and y = -1. These values also satisfy the thied equation. Hence, the given four points are coplanar.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
The probability that a person who undergoes a kidney operation will be recovered is $0.5.$ Find the probability that out of $6$ patients who undergo similar operation half of them recover.
→Prove by vector method that the angle subtended on semicircle is a right angle.
Evaluate the following integrals:$\int_{0}^\limits{1}\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)\text{dx}$
→In the figure 5.34 express $\bar{c}$ and $d$ in terms of $\bar{a}$ and $b$. Find a vector in the direction of $\bar{a}=$
→$\hat{i}-2 \hat{j}$ that has magnitude 7 units.
The probability of student A passing an examination is $\frac{2}{9}$ and of student B passing is $\frac{5}{9}$. Assuming the two events: 'A passes', 'B passes' as independent, find the probability of:
Only one of them passing the examination.
→Only one of them passing the examination.
Find the points o local maxima or local minima, if any, of the following functions, using the first derivatives test. Also, find the local maximum or local minimum values, as the case may be:
$\text{f}(\text{x})=\text{x}\sqrt{1-\text{x}}, \text{x}\geq0$
→$\text{f}(\text{x})=\text{x}\sqrt{1-\text{x}}, \text{x}\geq0$
Solve the following:
$\cos^{-1}\text{x}+\sin^{-1}\frac{\text{x}}{2}=\frac{\pi}{6}$
→$\cos^{-1}\text{x}+\sin^{-1}\frac{\text{x}}{2}=\frac{\pi}{6}$
Find the probability distribution of the number of doublets in three throws of a pair of dice
If $\text{y}=\text{e}^{-\text{x}}\cos\text{x},$ show that $\frac{\text{d}^2\text{y}}{\text{dx}^2}=2\text{e}^{-\text{x}}\sin\text{x}.$
→Transform $\left[\begin{array}{ccc}1 & 2 & 4 \\ 3 & -1 & 5 \\ 2 & 4 & 6\end{array}\right]$ into an upper triangular matrix by using suitable row transformations
→