The foot of perpendicular drawn from the point (0,0,0) to the plane is (4, -2, -5) then the equation of the plane is
- A4x + y + 5z = 14
- ✓4x – 2y – 5z = 45
- Cx – 2y – 5z = 10
- D4x + y + 6z = 11
Answer: B.
View full solution →388 questions across 7 question groups — pick any mix to generate a Maths paper with step-by-step answer keys.
MCQ
269 Q→02Answer the following questions in short.
10 Q→03Solve the Following Question.(2 Marks)
28 Q→04Solve the Following Question.(3 Marks)
42 Q→05Solve the Following Question.(4 Marks)
19 Q→06Do as instructed
1 Q→07Solve the Following Question.(5 Marks)
19 Q→One sample from each question group in this chapter. Select any group above to see the full set with answer keys.
Answer: B.
View full solution →m is:
Answer: A.
View full solution →Answer: A.
View full solution →Answer: B.
View full solution →Answer: A.
View full solution →plane $\bar{r} \cdot(6 \hat{i}+8 \hat{j}+7 \hat{k})=0$
$\bar{r}=(\hat{i}+4 \hat{j}+\hat{k})+\lambda(\hat{i}+2 \hat{j}+\hat{k})$.
$\bar{r}=(3 \hat{j}-\hat{k})+\mu(2 \hat{i}+3 \hat{j}+4 \hat{k})$
$\bar{r} \cdot(2 \hat{i}-\hat{j}+\hat{k})=23$
to $x-1=y-2=z-1$ and intersects the $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$.
$\bar{r}=(2 \hat{i}+6 \hat{j}+3 \hat{k})+\mu(2 \hat{i}+3 \hat{j}+4 \hat{k})$ are coplanar. Find the equation of the plane
determined by them.
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