1
TG EAPCET 2024 (Online) 9th May Evening Shift
MCQ (Single Correct Answer)
+1
-0
$r$ is a vector perpendicular to the planet, determined by the vectors $2 \hat{\mathbf{i}}-\hat{\mathbf{j}}$ and $\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$, If the magnitude of the projection of $\mathbf{r}$ on the vector $2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ is l , then $|\mathbf{r}|=$
A
$\sqrt{6}$
B
$3 \sqrt{6}$
C
$\frac{2 \sqrt{6}}{3}$
D
$\frac{3 \sqrt{6}}{2}$
2
TG EAPCET 2024 (Online) 9th May Evening Shift
MCQ (Single Correct Answer)
+1
-0
$\mathbf{b}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \mathbf{k}, \quad \mathbf{c}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}$ are two vectors and $\mathbf{a}$ is a vector such that $\cos (\mathbf{a}, \mathbf{b} \times \mathbf{c})=\sqrt{\frac{2}{3}}$. If $\mathbf{a}$ is a unit vector, then $|\mathbf{a} \times(\mathbf{b} \times \mathbf{c})|=$
A
3
B
2
C
1
D
4
3
TG EAPCET 2024 (Online) 9th May Evening Shift
MCQ (Single Correct Answer)
+1
-0
$A(3,2,-1), B(4,1,0), C(2,1,4)$ are the vertices of a $\triangle A B C$. If the bisector of $B A C$ ! intersects the side $B C$ at $D(p, q, r)$, then $\sqrt{2 p+q+r}=$
A
3
B
4
C
1
D
2
4
TG EAPCET 2024 (Online) 9th May Evening Shift
MCQ (Single Correct Answer)
+1
-0
$(3,0,2)$ and $(0,2, k)$ are the direction ratios of two lines and $\theta$ is the angle between them. If $|\cos \theta|=\frac{6}{13}$, then $k=$
A
$\pm 2$
B
$\pm 3$
C
$\pm 5$
D
$\pm 7$
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