Let O-A- i- 3j -2k and O-B- 3i- j -2k- then the vector O-C- bisecting the - AOB and C being a point on the line AB is

The correct option is A 2( i+ j k) In triangle the angle bisector divides the side A⃗B⃗ in equal ratio So point C is mid point #160; c⃗=O⃗A⃗ ...

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Applications of Dot Product

Let O⃗A⃗= i...

Question

Let $\to O A = ˆ i + 3 ˆ j - 2 ˆ k$ and $\to O B = 3 ˆ i + ˆ j - 2 ˆ k$ , then the vector $\to O C$ bisecting the $∠ A O B$ and $C$ being a point on the line $A B$ is

4 (ˆ i + ˆ j - ˆ k)

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2 (ˆ i + ˆ j - ˆ k)

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ˆ i + ˆ j - ˆ k

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None of these

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[\vec{a} \vec{b} \vec{c}]

\vec{a} = 2 \hat{i} - 3 \hat{j}, \vec{b} = \hat{i} + \hat{j} - \hat{k} and \vec{c} = 3 \hat{i} - \hat{k}

\vec{a} = \hat{i} - 2 \hat{j} + 3 \hat{k}, \vec{b} = 2 \hat{i} + \hat{j} - \hat{k} and \vec{c} = \hat{j} + \hat{k}

\vec{a} = 2 \hat{i} + 3 \hat{j} + \hat{k}, \vec{b} = \hat{i} - 2 \hat{j} + \hat{k} and \vec{c} = - 3 \hat{i} + \hat{j} + 2 \hat{k}

2 \hat{i} - \hat{j} + \hat{k}, \hat{i} - 3 \hat{j} - 5 \hat{k} and 3 \hat{i} - 4 \hat{j} - 4 \hat{k}

\hat{i} + \hat{j} + \hat{k}, 2 \hat{i} + 3 \hat{j} - \hat{k} and - \hat{i} - 2 \hat{j} + 2 \hat{k}

\to O A = i + j + k, \to A B = 3 i - 2 j + k, \to B C = i + 2 j - 2 k, \to C D = 2 i + j + 3 k

A unit vector in the plane of the vectors 2i + j + k, i - j + k and orthogonal to 5i + 2j + 6k is

\vec{a} and \vec{b}

\vec{a} = \hat{i} - \hat{j} and \vec{b} = \hat{j} + \hat{k}

\vec{a} = 3 \hat{i} - 2 \hat{j} - 6 \hat{k} and \vec{b} = 4 \hat{i} - \hat{j} + 8 \hat{k}

\vec{a} = 2 \hat{i} - \hat{j} + 2 \hat{k} and \vec{b} = 4 \hat{i} + 4 \hat{j} - 2 \hat{k}

\vec{a} = 2 \hat{i} - 3 \hat{j} + \hat{k} and \vec{b} = \hat{i} + \hat{j} - 2 \hat{k}

\vec{a} = \hat{i} + 2 \hat{j} - \hat{k}, \vec{b} = \hat{i} - \hat{j} + \hat{k}

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