1
BITSAT 2023
+3
-1

Let $$\mathbf{a}=2 \mathbf{i}+\mathbf{j}+\mathbf{k}, \mathbf{b}=\mathbf{i}+2 \mathbf{j}-\mathbf{k}$$ and $$a$$ unit vector $$\mathbf{c}$$ be coplanar. If $$\mathbf{c}$$ is perpendicular to $$\mathbf{a}$$, then c equals to

A
$$\frac{1}{\sqrt{5}}(\hat{\mathbf{i}}-2 \hat{\mathbf{j}})$$
B
$$\frac{1}{\sqrt{2}}(-\hat{\mathbf{j}}+\hat{\mathbf{k}})$$
C
$$\frac{1}{\sqrt{3}}(\hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}})$$
D
$$\frac{1}{\sqrt{3}}(-\hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}})$$
2
BITSAT 2022
+3
-1

$$\widehat u$$ and $$\widehat v$$ are two non-collinear unit vectors such that $$\left| {{{\widehat u + \widehat v} \over 2} + \widehat u \times \widehat v} \right| = 1$$. Then the value of $$|\widehat u \times \widehat v|$$ is equal to

A
$$\left| {{{\widehat u + \widehat v} \over 2}} \right|$$
B
$$|\widehat u + \widehat v|$$
C
$$|\widehat u - \widehat v|$$
D
$$\left| {{{\widehat u - \widehat v} \over 2}} \right|$$
3
BITSAT 2021
+3
-1

The points with position vectors $$10\widehat i + 3\widehat j$$, $$12\widehat i - 5\widehat j$$ and $$a\widehat i + 11\widehat j$$ are collinear, if a is

A
8
B
4
C
2
D
$${{82} \over 9}$$
4
BITSAT 2021
+3
-1

Let a, b, c be vectors of lengths 3, 4, 5 respectively and a be perpendicular to (b + c), b to (c + a) and c to (a + b), then the value of (a + b + c) is

A
2$$\sqrt5$$
B
2$$\sqrt2$$
C
10$$\sqrt5$$
D
5$$\sqrt2$$
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