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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

If the direction cosines of a line are $k, k, k$ , then:

(A) $k > 0$

(B) $0 < k < 1$

(C) $k = 1$

(D) $k = \frac{1}{\sqrt[]{3}}$ or - $\frac{1}{\sqrt[]{3}}$

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Asked by Payal Gupta

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Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} I f l, m, n a r e t h e d i r e c t i o n c o s i n e s o f a l i n e, t h e n \\ l^{2} + m^{2} + n^{2} = 1 \\ S o, k^{2} + k^{2} + k^{2} = 1 \\ \Rightarrow 3 k^{2} = 1 \Rightarrow k = \pm \frac{1}{\sqrt[]{3}} \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

The distance of the point $(α, β, γ)$ from the y-axis is

(A) $β$

(B) $γ$

(C) $\sqrt[]{β^{2} + γ^{2}}$

(D) $α^{2} + γ^{2}$

2 Views|Posted a year ago

Asked by Payal Gupta

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Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} T h e g i v e n point i s (α, β, γ) \\ A n y point o n y - a x i s = (0, β, 0) \\ ∴ Required distance = \sqrt[]{{(α - 0)}^{2} + {(β - β)}^{2} + {(δ - 0)}^{2}} \\ = \sqrt[]{α^{2} + γ^{2}} \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

If $l_{1}, m_{1}, n_{1}$ ; $l_{2}, m_{2}, n_{2}$ , and  $l_{3}, m_{3}, n_{3}$ are the direction cosines of three mutually perpendicular lines, prove that the line whose direction cosines are proportional to $l_{1} + l_{2} + l_{3}$ , $m_{1} + m_{2} + m_{3}$ , $n_{1} + n_{2} + n_{3}$ and  makes equal angles with them.

5 Views|Posted a year ago

Asked by Payal Gupta

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Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} L e t \vec{a}, \vec{b}, \vec{c} a n d \vec{d} a r e s u c h t h a t \\ \vec{a} = l_{1} \hat{i} + m_{1} \hat{j} + n_{1} \hat{k} \\ \vec{b} = l_{2} \hat{i} + m_{2} \hat{j} + n_{2} \hat{k} \\ \vec{c} = l_{3} \hat{i} + m_{3} \hat{j} + n_{3} \hat{k} \\ a n d \vec{d} = (l_{1} + l_{2} + l_{3}) \hat{i} + (m_{1} + m_{2} + m_{3}) \hat{j} + (n_{1} + n_{2} + n_{3}) \hat{k} \\ Since t h e g i v e n d' cosines a r e m u t u a l l y p e r p e n d i c u l a r t h e n \\ l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2} = 0 \\ l_{2} l_{3} + m_{2} m_{3} + n_{2} n_{3} = 0 \\ l_{1} l_{3} + m_{1} m_{3} + n_{1} n_{3} = 0 \\ L e t α, β a n d γ b e t h e a n g l e s b e t w e e n \vec{a} a n d \vec{d}, \vec{b} a n d \vec{d}, \vec{c} a n d \vec{d} r e s p e c t i v e l y . \\ ∴ c o s α = l_{1} (l_{1} + l_{2} + l_{3}) + m_{1} (m_{1} + m_{2} + m_{3}) + n_{1} (n_{1} + n_{2} + n_{3}) \\ = l_{1}^{2} + l_{1} l_{2} + l_{1} l_{3} + m_{1}^{2} + m_{1} m_{2} + m_{1} m_{3} + n_{1}^{2} + n_{1} n_{2} + n_{1} n_{3} \\ = (l_{1}^{2} + m_{1}^{2} + n_{1}^{2}) + (l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2}) + (l_{1} l_{3} + m_{1} m_{3} + n_{1} n_{3}) \\ = 1 + 0 + 0 = 1 \\ ∴ c o s β = l_{2} (l_{1} + l_{2} + l_{3}) + m_{2} (m_{1} + m_{2} + m_{3}) + n_{2} (n_{1} + n_{2} + n_{3}) \\ = l_{1} l_{2} + l_{2}^{2} + l_{2} l_{3} + m_{1} m_{2} + m_{2}^{2} + m_{2} m_{3} + n_{1} n_{2} + n_{2}^{2} + n_{2} n_{3} \\ = (l_{2}^{2} + m_{2}^{2} + n_{2}^{2}) + (l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2}) + (l_{2} l_{3} + m_{2} m_{3} + n_{2} n_{3}) \\ = 1 + 0 + 0 = 1 \\ S i m i l a r l y, \\ ∴ c o s γ = l_{3} (l_{1} + l_{2} + l_{3}) + m_{3} (m_{1} + m_{2} + m_{3}) + n_{3} (n_{1} + n_{2} + n_{3}) \\ = l_{1} l_{3} + l_{2} l_{3} + l_{3}^{2} + m_{1} m_{3} + m_{2} m_{3} + m_{3}^{2} + n_{1} n_{3} + n_{2} n_{3} + n_{3}^{2} \\ = (l_{3}^{2} + m_{3}^{2} + n_{3}^{2}) + (l_{1} l_{3} + m_{1} m_{3} + n_{1} n_{3}) + (l_{2} l_{3} + m_{2} m_{3} + n_{2} n_{3}) \\ = 1 + 0 + 0 = 1 \\ ∴ c o s α = c o s β = c o s γ = 1 \Rightarrow α = β = γ w h i c h i s t h e r e q u i r e d r e s u l t . \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

Show that the straight lines whose direction cosines are given by  $2 l + 2 m - n = 0 and m n + n l + l m = 0$ are at right angles.

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Asked by Payal Gupta

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Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t 2 l + 2 m - n = 0 \dots (i) \\ a n d m n + n l + l m = 0 \dots (i i) \\ Eliminating m f r o m e q . (i) a n d (i i) w e g e t, \\ m = \frac{n - 2 l}{2} [F r o m (i)] \\ \Rightarrow (\frac{n - 2 l}{2}) n + n l + l (\frac{n - 2 l}{2}) = 0 \\ \Rightarrow \frac{n^{2} - 2 n l + 2 n l + n l - 2 l^{2}}{2} = 0 \\ \Rightarrow n^{2} + n l - 2 l^{2} = 0 \\ \Rightarrow n^{2} + 2 n l - n l - 2 l^{2} = 0 \\ \Rightarrow n (n + 2 l) - l (n + 2 l) = 0 \\ \Rightarrow (n - l) (n + 2 l) = 0 \\ \Rightarrow n = - 2 l a n d n = l \\ ∴ m = \frac{- 2 l - 2 l}{2}, m = \frac{l - 2 l}{2} \\ \Rightarrow m = - 2 l, m = \frac{- l}{2} \\ T h e r e f o r e, t h e d i r e c t i o n r a t i o s a r e p r o p o r t i o n a l t o l, - 2 l, - 2 l a n d l, \frac{- l}{2}, l . \\ \Rightarrow 1, - 2, - 2 a n d 2, - 1, 2 \\ I f t h e t w o l i n e s a r e p e r p e n d i c u l a r t o e a c h o t h e r t h e n \\ 1 (2) - 2 (- 1) - 2 \times 2 = 0 \\ 2 + 2 - 4 = 0 \\ 0 = 0 \\ H e n c e, t h e t w o l i n e s a r e p e r p e n d i c u l a r . \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

AB = 3 $\hat{i}$ -  $\hat{j}$ + $\hat{k}$ and  $C D$ = -3 $\hat{i}$ + 2 $\hat{j}$ + 4 $\hat{k}$ are two vectors. The position vectors of the points A and C are 6 $\hat{i}$ + 7 $\hat{j}$  + 4 $\hat{k}$ and -9 $\hat{i}$  + 2 $\hat{j}$  + $\hat{k,}$ respectively. Find the position vector of a point P on the line AB and a point Q on the line CD such that $\vec{P Q}$  is perpendicular to both  $\vec{A B}$ and  $\vec{C D}$ both.

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Asked by Payal Gupta

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Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} P o s i t i o n v e c t o r o f A i s 6 \hat{i} + 7 \hat{j} + 4 \hat{k} a n d \vec{A B} = 3 \hat{i} - \hat{j} + \hat{k} \\ S o, e q u a t i o n o f a n y l i n e passing t h r o u g h A a n d p a r a l l e l t o \vec{A B} \\ \vec{r} = (6 \hat{i} + 7 \hat{j} + 4 \hat{k}) + λ (3 \hat{i} - \hat{j} + \hat{k}) \dots (i) \\ N o w a n y point P o n \vec{A B} = (6 + 3 λ, 7 - λ, 4 + λ) \\ S i m i l a r l y, p o s i t i o n v e c t o r o f C i s - 9 \hat{j} + 2 \hat{k} a n d \vec{C D} = - 3 \hat{i} + 2 \hat{j} + 4 \hat{k} \\ S o, e q u a t i o n o f a n y l i n e passing t h r o u g h C a n d p a r a l l e l t o \vec{C D} i s \\ \vec{r} = (- 9 \hat{j} + 2 \hat{k}) + μ (- 3 \hat{i} + 2 \hat{j} + 4 \hat{k}) \dots (i i) \\ A n y point Q o n \vec{C D} = (- 3 μ, - 9 + 2 μ, 2 + 4 μ) \\ d' r a t i o s o f \vec{P Q} a r e \\ (6 + 3 λ + 3 μ, 7 - λ + 9 - 2 μ, 4 + λ - 2 - 4 μ) \\ \Rightarrow (6 + 3 λ + 3 μ), (16 - λ - 2 μ), (2 + λ - 4 μ) \\ N o w \vec{P Q} i s ⊥ t o e q . (i), t h e n \\ 3 (6 + 3 λ + 3 μ) - 1 (16 - λ - 2 μ) + 1 (2 + λ - 4 μ) = 0 \\ \Rightarrow 18 + 9 λ + 9 μ - 16 + λ + 2 μ + 2 + λ - 4 μ = 0 \\ \Rightarrow 11 λ + 7 μ + 4 = 0 \dots (i i i) \\ \vec{P Q} i s ⊥ t o e q . (i i), t h e n \\ - 3 (6 + 3 λ + 3 μ) + 2 (16 - λ - 2 μ) + 4 (2 + λ - 4 μ) = 0 \\ \Rightarrow - 18 - 9 λ - 9 μ + 32 - 2 λ - 4 μ + 8 + 4 λ - 16 μ = 0 \\ \Rightarrow - 7 λ - 29 μ + 22 = 0 \\ \Rightarrow 7 λ + 29 μ - 22 = 0 \dots (i v) \\ S o l v i n g e q n . (i i i) a n d (i v) w e g e t \\ \Rightarrow 77 λ + 49 μ + 28 = 0 \\ 77 λ + 319 μ - 242 = 0 \\ \underline{(-) (-) (+)} \\ - 270 μ + 270 = 0 ∴ μ = 1 \\ P u t t i n g t h e v a l u e o f μ i n e q n . (i v) w e g e t, \\ 7 λ + 29 - 22 = 0 \Rightarrow λ = - 1 \\ ∴ P o s i t i o n v e c t o r o f P = [6 + 3 (- 1), 7 + 1, 4 - 1] = (3, 8, 3) \\ a n d P o s i t i o n v e c t o r o f Q = [- 3 (1), - 9 + 2 (1), 2 + 4 (1)] = (- 3, - 7, 6) \\ H e n c e, t h e p o s i t i o n v e c t o r o f \\ P = 3 \hat{i} + 8 \hat{j} + 3 \hat{k} a n d Q = - 3 \hat{i} - 7 \hat{j} + 6 \hat{k} \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

Show that the points $(\hat{i} - \hat{j} + \hat{k})$ and 3 $(\hat{i} + \hat{j} + \hat{k})$ are equidistant from the plane  and lie on opposite sides of it.

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Asked by Payal Gupta

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Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n points a r e P (\hat{i} - \hat{j} + 3 \hat{k}) a n d Q (3 \hat{i} + 3 \hat{j} + 3 \hat{k}) a n d t h e p l a n e \vec{r} . (5 \hat{i} + 2 \hat{j} - 7 \hat{k}) + 9 = 0 \\ P e r p e n d i c u l a r distance o f P (\hat{i} - \hat{j} + 3 \hat{k}) f r o m t h e p l a n e \\ \vec{r} . (5 \hat{i} + 2 \hat{j} - 7 \hat{k}) + 9 = | \frac{(\hat{i} - \hat{j} + 3 \hat{k}) . (5 \hat{i} + 2 \hat{j} - 7 \hat{k}) + 9}{\sqrt[]{{(5)}^{2} + {(2)}^{2} + {(- 7)}^{2}}} | \\ = | \frac{5 - 2 - 21 + 9}{\sqrt[]{25 + 4 + 49}} | = | \frac{- 9}{\sqrt[]{78}} | \\ a n d p e r p e n d i c u l a r distance o f Q (3 \hat{i} + 3 \hat{j} + 3 \hat{k}) f r o m t h e p l a n e \\ = | \frac{(3 \hat{i} + 3 \hat{j} + 3 \hat{k}) . (5 \hat{i} + 2 \hat{j} - 7 \hat{k}) + 9}{\sqrt[]{25 + 4 + 49}} | \\ = | \frac{15 + 6 - 21 + 9}{\sqrt[]{78}} | = | \frac{9}{\sqrt[]{78}} | \\ H e n c e, t h e t w o points a r e equidistant f r o m t h e g i v e n p l a n e . \\ O p p o s i t e s i g n s h o w s t h a t t h e y l i e o n e i t h e r s i d e o f t h e p l a n e . \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

Find the equation of the plane through the intersection of the planes  $\vec{r} \cdot (\hat{i} + 3 \hat{j}) - 6 = 0 and \vec{r} \cdot (3 \hat{i} - \hat{j} - 4 \hat{k}) = 0,$ whose perpendicular distance from the origin is unity.

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Asked by Payal Gupta

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Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n p l a n e s a r e : \\ \vec{r} . (\hat{i} - 3 \hat{j}) - 6 = 0 \Rightarrow x + 3 y - 6 = 0 \dots (i) \\ a n d \vec{r} . (3 \hat{i} - \hat{j} - 4 \hat{k}) = 0 \Rightarrow 3 x - y - 4 z = 0 \dots (i i) \\ E q u a t i o n o f t h e p l a n e passing t h r o u g h t h e l i n e o f intersection o f p l a n e (i) a n d (i i) i s \\ (x + 3 y - 6) + k (3 x - y - 4 z) = 0 \dots (i i i) \\ (1 + 3 k) x + (3 - k) y - 4 k z - 6 = 0 \\ P e r p e n d i c u l a r distance f r o m o r i g i n \\ \Rightarrow | \frac{- 6}{\sqrt[]{{(1 + 3 k)}^{2} + {(3 - k)}^{2} + {(- 4 k)}^{2}}} | = 1 \\ \Rightarrow \frac{36}{1 + 9 k^{2} + 6 k + 9 + k^{2} - 6 k + 16 k^{2}} = 1 [S q u a r i n g b o t h s i d e s] \\ \Rightarrow \frac{36}{26 k^{2} + 10} = 1 \Rightarrow 26 k^{2} + 10 = 36 \\ \Rightarrow 26 k^{2} = 26 \Rightarrow k^{2} = 1 ∴ k = \pm 1 \\ P u t t i n g t h e v a l u e o f k i n e q . (i i i) w e g e t, \\ (x + 3 y - 6) \pm (3 x - y - 4 z) = 0 \\ \Rightarrow x + 3 y - 6 + 3 x - y - 4 z = 0 a n d x + 3 y - 6 - 3 x + y + 4 z = 0 \\ \Rightarrow 4 x + 2 y - 4 z - 6 = 0 a n d - 2 x + 4 y + 4 z - 6 = 0 \\ H e n c e, t h e r e q u i r e d e q u a t i o n a r e \\ 4 x + 2 y - 4 z - 6 = 0 a n d - 2 x + 4 y + 4 z - 6 = 0 \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Ten

If $\vec{a} \cdot$ and $\vec{b}$ are adjacent sides of a rhombus, then $\vec{a} \cdot \vec{b}$ = 0.

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Asked by Alok Kumar Singh

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Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a True or False type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} I f \vec{a} . \vec{b} = 0 t h e n \vec{a} . \vec{b} = | \vec{a} | | \vec{b} | c o s 9 0^{0} \\ S o t h e a n g l e b e t w e e n t h e a d j a c e n t s i d e s o f t h e rhombus s h o u l d b e 9 0^{0} w h i c h i s n o t p o s s i b l e . \\ H e n c e t h e g i v e n s t a t e m e n t i s F a l s e . \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Ten

The formula ${\vec{(a}}^{} + \vec{b})^{2} + ({\vec{a}}^{2} + {\vec{b}}^{2}) + 2 {(\vec{a} \times \vec{b})}^{2}$ is valid for non-zero vectors $\vec{a}$ and $\vec{b}$ .

2 Views|Posted a year ago

Asked by Alok Kumar Singh

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Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a True or False type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} {(\vec{a} + \vec{b})}^{2} = (\vec{a} + \vec{b}) . (\vec{a} + \vec{b}) \\ = {| \vec{a} |}^{2} + {| \vec{b} |}^{2} + 2 \vec{a} . \vec{b} \\ H e n c e t h e g i v e n s t a t e m e n t i s F a l s e . \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

The plane  $a x + b y = 0$ is rotated about its line of intersection with the plane $z = 0$ through an angle $α$ . Prove that the equation of the plane in its new position is $a x + b y \pm \sqrt[]{a^{2} + b^{2}} t a n (α) z = 0 .$

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Asked by Payal Gupta

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Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} T h e g i v e n p l a n e s a r e \\ a x + b y = 0 \dots (i) \\ z = 0 \dots (i i) \\ E q u a t i o n o f a n y p l a n e passing t h r o u g h t h e l i n e o f intersection o f p l a n e (i) a n d (i i) i s \\ (a x + b y) + k z = 0 \Rightarrow a x + b y + k z = 0 \dots (i i i) \\ D i v i d i n g b o t h s i d e s b y \sqrt[]{a^{2} + b^{2} + k^{2}}, w e g e t \\ \frac{a}{\sqrt[]{a^{2} + b^{2} + k^{2}}} x + \frac{b}{\sqrt[]{a^{2} + b^{2} + k^{2}}} y + \frac{k}{\sqrt[]{a^{2} + b^{2} + k^{2}}} z = 0 \\ ∴ D i r e c t i o n cosines o f t h e n o r m a l t o t h e p l a n e a r e \\ \frac{a}{\sqrt[]{a^{2} + b^{2} + k^{2}}}, \frac{b}{\sqrt[]{a^{2} + b^{2} + k^{2}}}, \frac{k}{\sqrt[]{a^{2} + b^{2} + k^{2}}} \\ a n d t h e d i r e c t i o n cosines o f t h e p l a n e (i) a r e \\ \frac{a}{\sqrt[]{a^{2} + b^{2}}}, \frac{b}{\sqrt[]{a^{2} + b^{2}}}, 0 \\ Since, α i s t h e a n g l e b e t w e e n t h e p l a n e s (i) a n d (i i i), w e g e t \\ c o s α = \frac{a . a + b . b + k . 0}{\sqrt[]{a^{2} + b^{2} + k^{2}} . \sqrt[]{a^{2} + b^{2}}} \\ \Rightarrow c o s α = \frac{a^{2} + b^{2}}{\sqrt[]{a^{2} + b^{2} + k^{2}} . \sqrt[]{a^{2} + b^{2}}} \\ \Rightarrow c o s α = \frac{\sqrt[]{a^{2} + b^{2}}}{\sqrt[]{a^{2} + b^{2} + k^{2}}} \Rightarrow c o s^{2} α = \frac{a^{2} + b^{2}}{a^{2} + b^{2} + k^{2}} \\ \Rightarrow (a^{2} + b^{2} + k^{2}) c o s^{2} α = a^{2} + b^{2} \\ \Rightarrow a^{2} c o s^{2} α + b^{2} c o s^{2} α + k^{2} c o s^{2} α = a^{2} + b^{2} \\ \Rightarrow k^{2} c o s^{2} α = a^{2} - a^{2} c o s^{2} α + b^{2} - b^{2} c o s^{2} α \\ \Rightarrow k^{2} c o s^{2} α = a^{2} (1 - c o s^{2} α) + b^{2} (1 - c o s^{2} α) \\ \Rightarrow k^{2} c o s^{2} α = a^{2} s i n^{2} α + b^{2} s i n^{2} α \\ \Rightarrow k^{2} c o s^{2} α = (a^{2} + b^{2}) s i n^{2} α \\ \Rightarrow k^{2} = (a^{2} + b^{2}) \frac{s i n^{2} α}{c o s^{2} α} \Rightarrow k = \pm \sqrt[]{a^{2} + b^{2}} . t a n α \\ P u t t i n g t h e v a l u e o f k i n e q . (i i i) w e g e t \\ a x + b y \pm (\sqrt[]{a^{2} + b^{2}} . t a n α) z = 0 w h i c h i s t h e r e q u i r e d E q u a t i o n o f p l a n e . \\ H e n c e p r o v e d . \end{array}$

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If the direction cosines of a line are $k, k, k$ , then:

(A) $k > 0$

(B) $0 < k < 1$

(C) $k = 1$

(D) $k = \frac{1}{\sqrt[]{3}}$ or - $\frac{1}{\sqrt[]{3}}$

The distance of the point $(α, β, γ)$ from the y-axis is

(A) $β$

(B) $γ$

(C) $\sqrt[]{β^{2} + γ^{2}}$

(D) $α^{2} + γ^{2}$

Show that the straight lines whose direction cosines are given by  $2 l + 2 m - n = 0 and m n + n l + l m = 0$ are at right angles.

Show that the points $(\hat{i} - \hat{j} + \hat{k})$ and 3 $(\hat{i} + \hat{j} + \hat{k})$ are equidistant from the plane  and lie on opposite sides of it.

Find the equation of the plane through the intersection of the planes  $\vec{r} \cdot (\hat{i} + 3 \hat{j}) - 6 = 0 and \vec{r} \cdot (3 \hat{i} - \hat{j} - 4 \hat{k}) = 0,$ whose perpendicular distance from the origin is unity.

If $\vec{a} \cdot$ and $\vec{b}$ are adjacent sides of a rhombus, then $\vec{a} \cdot \vec{b}$ = 0.

The formula ${\vec{(a}}^{} + \vec{b})^{2} + ({\vec{a}}^{2} + {\vec{b}}^{2}) + 2 {(\vec{a} \times \vec{b})}^{2}$ is valid for non-zero vectors $\vec{a}$ and $\vec{b}$ .