The vector equation of the straight line \frac{1-x}{3}=\frac{y+1}{-2}=\frac{3-z}{-1} is

Correct Answer is : →r=(∧i−∧j+3∧k)+λ(3∧i+2∧j−∧k)

\overrightarrow{r}=(\hat{i}-\hat{j}+3 \hat{k})+\lambda(3 \hat{i}+2 \hat{j}-\hat{k})

\overrightarrow{r}=(\hat{i}-\hat{j}+3 \hat{k})+\lambda(3 \hat{i}-2 \hat{j}-\hat{k})

\overrightarrow{r}=(3 \hat{i}-2 \hat{j}-\hat{k})+\lambda(\hat{i}-\hat{j}+3 \hat{k})

\overrightarrow{r}=(3 \hat{i}+2 \hat{j}-\hat{k})+\lambda(\hat{i}-\hat{j}+3 \hat{k})

\overrightarrow{r}=(\hat{i}-\hat{j}+3 \hat{k})+\lambda(3 \hat{i}+2 \hat{j}+\hat{k})

KEAM 2010 Math Paper

Question : 19 of 120

Marks: +1, -0

The vector equation of the straight line ‌

1−x

=‌

y+1

−2

=‌

3−z

−1

→
r
=(
∧
i
−
∧
j
+3
∧
k
)+λ(3
∧
i
+2
∧
j
−
∧
k
)
→
r
=(
∧
i
−
∧
j
+3
∧
k
)+λ(3
∧
i
−2
∧
j
−
∧
k
)
→
r
=(3
∧
i
−2
∧
j
−
∧
k
)+λ(
∧
i
−
∧
j
+3
∧
k
)
→
r
=(3
∧
i
+2
∧
j
−
∧
k
)+λ(
∧
i
−
∧
j
+3
∧
k
)
→
r
=(
∧
i
−
∧
j
+3
∧
k
)+λ(3
∧
i
+2
∧
j
+
∧
k
)

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