Find the angle(in degrees) between the vectors $a = 2 i + 5 j - k a n d b = i - j - 3 k .$

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Solution

Here $\to a = 2^i + 5^j -^k$ and $\to b =^i -^j - 3^k$
So, $\to a \cdot \to b = (2^i + 5^j -^k) . (^i -^j - 3^k)$
$= 2 (1) + 5 (- 1) + (- 1) (- 3)$
$= 2 - 5 + 3$
$= 0$
$\to a \cdot \to b = 0$ means both vectors $\to a$ and $\to b$ are perpendicular hence angle between them is $90^{\circ}$ .

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Similar questions

Q. If

\to a = - \to i + \to j + \to k and \to b = 2 \to i + \to j + \to k,

then the vector

\to c

satisfying the conditions

1) coplanar with \to a and \to b

2) perpendicular to \to b

3) \to a \cdot \to c = 7

The position vectors of four [point A, B, C and D are $ˆ i + ˆ j + ˆ k, 2 ˆ i + 5 ˆ j, 3 ˆ i + 2 ˆ j - 3 ˆ k, ˆ i - 6 ˆ j - ˆ k$ then select the correct relation:

Q. (i) Dot product of a vector with

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

are 0, 5 and 8 respectively. Find the vector.
(ii) Dot products of a vector with vectors

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

are respectively 4, 0 and 2. Find the vector.

Q. The distance between the plane whose equation is

¯ ¯ ¯ r \cdot (2 ¯ i + ¯ j - 3 ¯ ¯ ¯ k) = 5

and the line whose equation is

¯ ¯ ¯ r = i + λ (2 ¯ i + 5 ¯ j + 3 ¯ ¯ ¯ k)

Q. Show that the vectors

\vec{a,} \vec{b,} \vec{c}

given by

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

are non-coplanar. Express vector

\vec{d} = 2 \hat{i} - \hat{j} - 3 \hat{k}

as a linear combination of the vectors

\vec{a,} \vec{b} and \vec{c} .

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Find the angle(in degrees) between the vectors a=2i+5j−kandb=i−j−3k.

Here →a=2^i+5^j−^k and →b=^i−^j−3^kSo, →a⋅→b=(2^i+5^j−^k).(^i−^j−3^k)=2(1)+5(−1)+(−1)(−3)=2−5+3=0→a⋅→b=0 means both vectors →a and →b are perpendicular hence angle between them is 90∘.

Find the angle(in degrees) between the vectors $a = 2 i + 5 j - k a n d b = i - j - 3 k .$

Here $\to a = 2^i + 5^j -^k$ and $\to b =^i -^j - 3^k$
So, $\to a \cdot \to b = (2^i + 5^j -^k) . (^i -^j - 3^k)$
$= 2 (1) + 5 (- 1) + (- 1) (- 3)$
$= 2 - 5 + 3$
$= 0$
$\to a \cdot \to b = 0$ means both vectors $\to a$ and $\to b$ are perpendicular hence angle between them is $90^{\circ}$ .